Topology Course Lecture Notes

Aisling McCluskey and Brian McMaster

August 1997

Chapter 1
Fundamental Concepts

• many of the concepts can be described purely in terms of open sets,
• open-set descriptions are sometimes simpler than metric descriptions, e.g. continuity,
• many results about these concepts can be proved using only the basic properties of open sets (namely, that both the empty set and the underlying set X are open, that the intersection of any two open sets is again open and that the union of arbitrarily many open sets is open).

This prompts the question: How far would we get if we started with a collection of subsets possessing these above-mentioned properties and proceeded to define everything in terms of them?

1.1Describing Topological Spaces

• the empty set and underlying set X are both open,
• the intersection of any two open sets is open, and
• unions of arbitrarily many open sets are open.

We will call any collection of sets on X satisfying these properties a topology. In the following section, we also seek to give alternative ways of describing this important collection of sets.

1.1.1 Defining Topological Spaces

Examples

(i) Any metric space (X,d) is a topological space where T_d, the topology for X induced by the metric d, is defined by agreeing that G shall be declared as open whenever each x in G is contained in an open ball entirely in G, i.e.

Æ Ì G Í X is open in (X, Td) Û "x Î G, $rx > 0 such that x Î Brx(x) Í G.


(ii) The following is a special case of (i), above. Let R be the set of real numbers and let Á be the usual (metric) topology defined by agreeing that

Æ Ì G Í X is open in (R, Á) (alternatively, Á-open) Û "x Î G, $rx > 0 such that (x-rx, x+rx) Ì G.


(iii) Define T₀ = {Æ, X}
(iv) Define D = { G Í X: G Í X }
(v) For any non-empty set X, the family C = { G Í X: G = Æ or X\G is finite} is a topology for X called the cofinite topology.
(vi) For any non-empty set X, the family L = { G Í X: G = Æ or X\G is countable} is a topology for X called the cocountable topology.

1.1.2 Neighbourhoods

Occasionally, arguments can be simplified when the sets involved are not ``over-described''. In particular, it is sometimes suffices to use sets which contain open sets but are not necessarily open. We call such sets neighborhoods.

Definition 2 Given a topological space (X, T) with x Î X, then N Í X is said to be a (T)-neighbourhood of x Û $ open set G with x Î G Í N.

It follows then that a set U Í X is open iff for every x Î U, there exists a neighbourhood N_x of x contained in U. (Check this!)

Lemma 1 Let (X, T) be a topological space and, for each x Î X, let N(x) be the family of neighbourhoods of x. Then

(i) U Î N(x) Þ x Î U.
(ii) N(x) is closed under finite intersections.
(iii) U Î N(x) and U Í V Þ V Î N(x).
(iv) U Î N(x) Þ $ W Î N(x) such that W Í U and W Î N(y) for each y Î W.

Proof Exercise!

Examples

(i) Let x Î X, and define T_x = { Æ, {x}, X }. Then T_x is a topology for X and V Í X is a neighbourhood of x iff x Î V. However, the only nhd of y Î X where y ¹ x is X itself
(ii) Let x Î X and define a topology Á(x) for X as follows:

Á(x) = { G Í X: x Î G} È{ Æ}.

Note here that every nhd of a point in X is open.
(iii) Let x Î X and define a topology E(x) for X as follows:

E(x) = { G Í X: x \not Î G } È{X}.

Note here that {y} is open for every y ¹ x in X, that {x,y} is not open, is not a nhd of x yet is a nhd of y.

In fact, the only nhd of x is X.

1.1.3 Bases and Subbases

It often happens that the open sets of a space can be very complicated and yet they can all be described using a selection of fairly simple special ones. When this happens, the set of simple open sets is called a base or subbase (depending on how the description is to be done). In addition, it is fortunate that many topological concepts can be characterized in terms of these simpler base or subbase elements.

Definition 3 Let (X,T) be a topological space. A family B Í T is called a base for (X,T) if and only if every non-empty open subset of X can be represented as a union of a subfamily of B.

It is easily verified that B Í T is a base for (X, T) if and only if whenever x Î G Î T, $B Î B such that x Î B Í G.

Clearly, a topological space can have many bases.

Lemma 2 If B is a family of subsets of a set X such that

(B1) for any B₁, B₂ Î B and every point x Î B₁ ÇB₂, there exists B₃ Î B with x Î B₃ Í B₁ ÇB₂, and
(B2) for every x Î X, there exists B Î B such that x Î B,

Conversely, any base B for a topological space (X,T) satisfies (B1) and (B2).

Proof (Exercise!)

Definition 4 Let (X,T) be a topological space. A family S Í T is called a subbase for (X,T) if and only if the family of all finite intersections Ç_{i = 1}^kU_i, where U_i Î S for i = 1,2,¼,k is a base for (X,T).

Examples

(i) In any metric space (X,d), { B_r(x):x Î X, r > 0 } forms a base for the induced metric topology T_d on X.
(ii) For the real line R with its usual (Euclidean) topology, the family {(a,b): a,b Î Q, a < b } is a base.
(iii) For an arbitrary set X, the family { {x}:x Î X } is a base for (X, D).
(iv) The family of all `semi-infinite' open intervals (a, ¥) and (-¥,b) in R is a subbase for (R, Á).

1.1.4 Generating Topologies

From the above examples, it follows that for a set X one can select in many different ways a family T such that (X,T) is a topological space. If T₁ and T₂ are two topologies for X and T₂ Í T₁, then we say that the topology T₁ is finer than the topology T₂, or that T₂ is coarser than the topology T₁. The discrete topology for X is the finest one; the trivial topology is the coarsest.

If X is an arbitrary infinite set with distinct points x and y, then one can readily verify that the topologies Á(x) and Á(y) are incomparable i.e. neither is finer than the other.

By generating a topology for X, we mean selecting a family T of subsets of X which satisfies conditions (T1)-(T3). Often it is more convenient not to describe the family T of open sets directly. The concept of a base offers an alternative method of generating topologies.

Examples

• [Sorgenfrey line] Given the real numbers R, let B be the family of all intervals [x,r) where x,r Î R, x < r and r is rational. One can readily check that B has properties (B1)-(B2). The space R_s, generated by B, is called the Sorgenfrey line and has many interesting properties. Note that the Sorgenfrey topology is finer than the Euclidean topology on R. (Check!)
• [Niemytzki plane] Let L denote the closed upper half-plane. We define a topology for L by declaring the basic open sets to be the following:

  
  (I) the (Euclidean) open discs in the upper half-plane;
  (II) the (Euclidean) open discs tangent to the `edge' of the L, together with the point of tangency.

  Note If y_n ® y in L, then

  
  (i) y not on `edge': same as Euclidean convergence.
  (ii) y on the `edge': same as Euclidean, but y_n must approach y from `inside'. Thus, for example, y_n = ([1/n],0) \not ® (0,0)!

1.1.5 New Spaces from Old

A subset of a topological space inherits a topology of its own, in an obvious way:

Definition 5 Given a topological space (X, T) with A Í X, then the family T_A = {A ÇG:G Î T } is a topology for A, called the subspace (or relative or induced) topology for A. (A, T_A) is called a subspace of (X, T).

Example
The interval I = [0,1] with its natural (Euclidean) topology is a (closed) subspace of (R, Á).

Warning: Although this definition, and several of the results which flow from it, may suggest that subspaces in general topology are going to be `easy' in the sense that a lot of the structure just gets traced onto the subset, there is unfortunately a rich source of mistakes here also: because we are handling two topologies at once. When we inspect a subset B of A, and refer to it as 'open' (or 'closed', or a 'neighbourhood' of some point p .... ) we must be exceedingly careful as to which topology is intended. For instance, in the previous example, [0,1] itself is open in the subspace topology on I but, of course, not in the 'background' topology of R. In such circumstances, it is advisable to specify the topology being used each time by saying T-open, T_A-open, and so on.

1.2 Closed sets and Closure

• define closed sets in a general topological space and
• examine the related notion of the closure of a given set.

1.2.1 Closed Sets

Definition 6 Given a topological space (X,T) with F Í X, then F is said to be T-closed iff its complement X \F is T-open.

From De Morgan's Laws and properties (T1)-(T3) of open sets, we infer that the family F of closed sets of a space has the following properties:

(F1) X Î F and Æ Î F
(F2) F is closed under finite union
(F3) F is closed under arbitrary intersection.

Sets which are simultaneously closed and open in a topological space are sometimes referred to as clopen sets. For example, members of the base B = { [x,r):x, r Î R, x < r, r rational } for the Sorgenfrey line are clopen with respect to the topology generated by B. Indeed, for the discrete space (X, D), every subset is clopen.

1.2.2 Closure of Sets

Definition 7 If (X,T) is a topological space and A Í X, then

Evidently, [`A]<sup>T</sup> (or [`A] when there is no danger of ambiguity) is the smallest closed subset of X which contains A. Note that A is closed Û A = [`A].

Lemma 3 If (X, T) is a topological space with A, B Í X, then

(i) [`(Æ)] = Æ
(ii) A Í [`A]
(iii) [`([`A])] = [`A]
(iv) [`(AÈB)] = [`A] È[`B].

Proof Exercise!

Theorem 1 Given a topological space with A Í X, then x Î [`A] iff for each nhd U of x, U ÇA ¹ Æ.

Proof

$\Rightarrow$: Let x Î [`A] and let U be a nhd of x; then there exists open G with x Î G Í U. If U ÇA = Æ, then G ÇA = Æ and so A Í X \G Þ [`A] Í X \G whence x Î X \G, thereby contradicting the assumption that U ÇA = Æ.
$\Leftarrow$: If x Î X \[`A], then X \[`A] is an open nhd of x so that, by hypothesis, (X \[`A]) ÇA ¹ Æ, which is a contradiction (i.e., a false statement).

Examples

(i) For an arbitrary infinite set X with the cofinite topology C, the closed sets are just the finite ones together with X. So for any A Í X,

_ A = ì í î A if A is finite X if A is infinite

Note that any two non-empty open subsets of X have non-empty intersection.


(ii) For an arbitrary uncountable set X with the cocountable topology L, the closed sets are the countable ones and X itself. Note that if we let X = R, then [`[0,1]] = R! (In the usual Euclidean topology, [`[0,1]] = [0,1].)
(iii) For the space (X, T_x) defined earlier, if Æ Ì A Í X, then

_ A = ì í î X if x Î A X \{x} if x \not Î A


(iv) For (X, Á(x)) with A Í X,

_ A = ì í î A if x \not Î A X if x Î A


(v) For (X, E(x)) with Æ Ì A Í X,

_ A = ì í î A if x Î A AÈ{x} if x \not Î A


(vi) In (X, D), every subset equals its own closure.

1.3 Continuity and Homeomorphism

Because many properties of spaces are preserved by continuous functions, spaces related by a bijection (one-to-one and onto function) which is continuous in both directions will have many properties in common. These properties are identified as topological properties. Spaces so related are called homeomorphic.

1.3.1 Continuity

Definition 8 Let (X, T) and (Y, S) be topological spaces; a mapping f:X® Y is called continuous iff f^-1(U) Î T for each U Î S i.e. the inverse image of any open subset of Y is open in X.

Examples

(i) If (X, D) is discrete and (Y, S) is an arbitrary topological space, then any function f:X ® Y is continuous!

Again, if (X, T) is an arbitrary topological space and (Y, T₀) is trivial, any mapping g:X ® Y is continuous.
(ii) If (X, T), (Y, S) are arbitrary topological spaces and f:X ® Y is a constant map, then f is continuous.
(iii) Let X be an arbitrary set having more than two elements, with x Î X. Let T = I(x), S = T_x in the definition of continuity; then the identity map id_X:X ® X is continuous. However, if we interchange T with S so that T = T_x and S = Á(x), then id_X:X ® X is not continuous! Note that id_X:(X, T₁) ® (X, T₂) is continuous if and only if T₁ is finer than T₂.

Theorem 2 If (X₁, T₁), (X₂, T₂) and (X₃, T₃) are topological spaces and h:X₁ ® X₂ and g:X₂ ® X₃ are continuous, then g°h:X₁ ® X₃ is continuous.

Proof Immediate.

There are several different ways to 'recognise' continuity for a mapping between topological spaces, of which the next theorem indicates two of the most useful apart from the definition itself:

Theorem 3 Let f be a mapping from a topological space (X₁, T₁) to a topological space (X₂, T₂). The following statements are equivalent:

(i) f is continuous,
(ii) the preimage under f of each closed subset of X₂ is closed in X₁,
(iii) for every subset A of X₁, f([`A]) Í [`f(A)].

Proof It is easy to see that (i) implies (ii). Assuming that (ii) holds, apply it to the closed set [`f(A)] and (iii) readily follows. Now if (iii) is assumed and G is a given open subset of X₂, use (iii) on the set A = X₁ \f^-1(G) and verify that it follows that f^-1(G) must be open.

1.3.2 Homeomorphism

Definition 9 Let (X, T), (Y, S) be topological spaces and let h:X ® Y be bijective. Then h is a homeomorphism iff h is continuous and h^-1 is continuous. If such a map exists, (X, T) and (Y, S) are called homeomorphic.

Such a map has the property that

Example
For every space (X, T), the identity mapping id_X:X ® X is a homeomorphism.

A property of topological spaces which when possessed by a space is also possessed by every space homeomorphic to it is called a topological invariant. We shall meet some examples of such properties later.

One can readily verify that if f is a homeomorphism, then the inverse mapping f^-1 is also a homeomorphism and that the composition g °f of two homeomorphisms f and g is again a homeomorphism. Thus, the relation `X and Y are homeomorphic' is an equivalence relation.

In general, it may be quite difficult to demonstrate that two spaces are homeomorphic (unless a homeomorphism is obvious or can easily be discovered). For example, to verify that (R, Á) is homeomorphic to (0,1) with its induced metric topology, it is necessary to demonstrate, for instance, that h:(0,1) ® R where h(x) = [(2x-1)/(x(x-1))] is a homeomorphism.

It is often easier to show that two spaces are not homeomorphic: simply exhibit an invariant which is possessed by one space and not the other.

Example
The spaces (X, Á(x)) and (X, E(x)) are not homeomorphic since, for example, (X, Á(x)) has the topological invariant `each nhd is open' while (X, E(x)) does not.

1.4 Additional Observations

Definition 10 A sequence (x_n) in a topological space (X, T) is said to converge to a point x Î X iff (x_n) eventually belongs to every nhd of x i.e. iff for every nhd U of x, there exists n₀ Î N such that x_n Î U for all n ³ n₀.

Caution

We learnt that, for metric spaces, sequential convergence was adequate to describe the topology of such spaces (in the sense that the basic primitives of `open set', `neighbourhood', `closure' etc. could be fully characterised in terms of sequential convergence). However, for general topological spaces, sequential convergence fails. We illustrate:

(i) Limits are not always unique. For example, in (X, T₀), each sequence (x_n) converges to every x Î X.
(ii) In R with the cocountable topology L, [0,1] is not closed and so G = (-¥,0) È(1,¥) is not open - yet if x_n ® x where x Î G, then Assignment 1 shows that x_n Î G for all sufficiently large n. Further, 2 Î [`[0,1]]^L, yet no sequence in [0,1] can approach 2. So another characterisation fails to carry over from metric space theory.

Finally, every L-convergent sequence of points in [0,1] must have its limit in [0,1] - but [0,1] is not closed (in L)!

Hence, to discuss topological convergence thoroughly, we need to develop a new basic set-theoretic tool which generalises the notion of sequence. It is called a net - we shall return to this later.

Definition 11 A topological space (X, T) is called metrizable iff there exists a metric d on X such that the topology T_d induced by d coincides with the original topology T on X.

The investigations above show that (X, T₀) and (R, L) are examples of non-metrizable spaces. However, the discrete space (X,D) is metrizable, being induced by the discrete metric

Chapter 2
Topological Properties

We explained in the previous chapter what a topological property (homeomorphic invariant) is but gave few good examples. We now explore some of the most important ones. Recurring themes will be:

• When do subspaces inherit the property?
• How do continuous maps relate to the property?
• Does the property behave specially in metric spaces?

2.1 Compactness

2.1.1 Compactness Defined

If A is a subspace of a space (X, T), U is an open cover for A iff U is a cover for A such that each member of U is open in X.

The classic theorem of Heine-Borel-Lebesgue asserts that, in R, every open cover of a closed bounded subset has a finite subcover. This theorem has extraordinarily profound consequences and like most good theorems, its conclusion has become a definition.

Definition 1 (X, T) is said to be compact iff every open cover of X has a finite subcover.

Theorem 1 [Alexander's Subbase Theorem] Let S be any subbase for (X,T). If every open cover of X by members of S has a finite subcover, then X is compact.

The proof of this deep result is an application of Zorn's lemma, and is not an exercise for the faint-hearted!

Examples

(i) (R, Á) is not compact, for consider U = { (-n,n): n Î N }. Similarly, (C, T_usual) is not compact.

(ii) (0,1) is not compact, for consider U = { ([1/n],1) :n ³ 2 }.

(iii) (X, C) is compact, for any X.

(iv) Given x Î X, (X, E(x)) is compact; (X, I(x)) is not compact unless X is finite.

(v) T finite for any X Þ (X, T) compact.

(vi) X finite, T any topology for X Þ (X, T) compact.

(vii) X infinite Þ (X, D) not compact.

(viii) Given (X, T), if (x_n) is a sequence in X convergent to x, then { x_n:n Î N } È{x} is compact.

2.1.2 Compactness for Subspaces

We call a subset A of (X,T) a compact subset when the subspace (A, T_A) is a compact space. It's a nuisance to have to look at T_A in order to decide on this. It would be easier to use the original T. Thankfully, we can!

Lemma 1 A is a compact subset of (X,T) iff every T-open cover of A has a finite subcover.

Proof Exercise.

Lemma 2 Compactness is closed-hereditary and preserved by continuous maps.

Proof Exercise.

Example
The unit circle in R² is compact; indeed, paths in any space are compact.

2.1.3 Compactness in Metric Spaces

In any metric space (M, d), every compact subset K is closed and bounded:
(bounded, since given any x₀ Î M,

Neither half is valid in all topological spaces;

• `compact Þ bounded' doesn't even make sense since `bounded' depends on the metric.
  
• `compact Þ closed' makes sense but is not always true. For example, in (R,C), (0,1) is not closed yet it is compact (since its topology is the cofinite topology!)

Further, in a metric space, a closed bounded subset needn't be compact (e.g. consider M with the discrete metric and let A Í M be infinite; then A is closed, bounded (since A Í B(x,2) = M for any x Î M), yet it is certainly not compact! Alternatively, the subspace (0,1) is closed (in itself), bounded, but not compact.)

However, the Heine-Borel theorem asserts that such is the case for R and Rⁿ; the following is a special case of the theorem:

Theorem 2 Every closed, bounded interval [a,b] in R is compact.

Proof Let U be any open cover of [a,b] and let K = { x Î [a,b]: [a,x] is covered by a finite subfamily of U}. Note that if x Î K and a £ y £ x, then y Î K. Clearly, K ¹ Æ since a Î K. Moreover, given x Î K, there exists d_x > 0 such that [x,x+ d_x) Í K (since x Î some open U Î chosen finite subcover of U). Since K is bounded, k^* = supK exists.

(i)	k* Î K	: Choose U Î U such that k* Î U; then there exists e > 0 such that (k*-e,k*] Í U. Since there exists x Î K such that k*-e < x < k*, k* Î K.
(ii)	k* = b	: If k* < b, choose U Î U with k* Î U and note that [k*,k*+d) Í U for some d > 0 -contradiction!

Note An alternative proof [Willard, Page 116] is to invoke the connected nature of [a,b] by showing K is clopen in [a,b].

Theorem 3 Any continuous map from a compact space into a metric space is bounded.

Proof Immediate.

Corollary 1 If (X, T) is compact and f:X ® R is continuous, then f is bounded and attains its bounds.

Proof Clearly, f is bounded. Let m = supf(X) and l = inff(X); we must prove that m Î f(X) and l Î f(X). Suppose that m \not Î f(X). Since [`f(X)] = f(X), then there exists e > 0 such that (m-e,m+e) Çf(X) = Æ i.e. for all x Î X, f(x) £ m-e¼contra!

Similarly, if l \not Î f(X), then there exists e > 0 such that [l,l+e) Çf(X) = Æ whence l+e is a lower bound for f(X)!

2.1.4 Sequential Compactness

Definition 2 A topological space (X, T) is said to be sequentially compact if and only if every sequence in X has a convergent subsequence.

Recall from Chapter 1 the definition of convergence of sequences in topological spaces and the cautionary remarks accompanying it. There we noted that, contrary to the metric space situation, sequences in topology can have several different limits! Consider, for example, (X,T₀) and (R,L). In the latter space, if x_n ® l, then x_n = l for all n ³ some n₀. Thus the sequence 1,[1/2], [1/4], [1/8], ¼ does not converge in (R,L)!

Lemma 3 Sequential compactness is closed-hereditary and preserved by continuous maps.

Proof Exercise.

We shall prove in the next section that in metric spaces, sequential compactness and compactness are equivalent!

Definition 3 Given a topological space (X,T), a subset A of X and x Î X, x is said to be an accumulation point of A iff every neighbourhood of x contains infinitely many points of A.

Lemma 4 Given a compact space (X,T) with an infinite subset A of X, then A has an accumulation point.

Proof Suppose not; then for each x Î X, there exists a neighbourhood N_x of x such that N_x ÇA is (at most) finite; the family { N_x:x Î X } is an open cover of X and so has a finite subcover { N_xi:i = 1, ¼,n }. But A Í X and A is infinite, whence

Lemma 5 Given a sequentially compact metric space (M,d) and e > 0, there is a finite number of open balls, radius e, which cover M.

Proof Suppose not and that for some e > 0, there exists no finite family of open balls, radius e, covering M. We derive a contradiction by constructing a sequence (x_n) inductively such that d(x_m,x_n) ³ e for all n, m (n ¹ m), whence no subsequence is even Cauchy!

Let x₁ Î M and suppose inductively that x₁, ¼,x_k have been chosen in M such that d(x_i,x_j) ³ e for all i, j £ k,i ¹ j.By hypothesis, {B(x_i,e):i = 1,¼,k } is not an (open) cover of M and so there exists x_k+1 Î M such that d(x_k+1,x_i) ³ e for 1 £ i £ k. We thus construct the required sequence (x_n), which clearly has no convergent subsequence.

Theorem 4 A metric space is compact iff it is sequentially compact.

Proof

$\Rightarrow :$ Suppose (M,d) is compact. Given any sequence (x_n) in M, either A = {x₁,x₂, ¼} is finite or it is infinite. If A is finite, there must be at least one point l in A which occurs infinitely often in the sequence and its occurrences form a subsequence converging to l. If A is infinite, then by the previous lemma there exists x Î X such that every neighbourhood of x contains infinitely many points of A.

For each k Î w, B(x,[1/k]) contains infinitely many x_n's: select one, call it x_nk, making sure that n_k > n_k-1 > n_k-2 ¼. We have a subsequence (x_n1, x_n2, ¼, x_nk, ¼) so that d(x,x_nk) < [1/k] ® 0 i.e. x_nk ® x. Thus in either case there exists a convergent subsequence and so (M,d) is sequentially compact.
$\Leftarrow :$ Conversely, suppose (M,d) is sequentially compact and not compact. Then there exists some open cover {G_i:i Î I} of M having no finite subcover. By Lemma 2.1.4, with e = [1/n] (n Î w), we can cover M by a finite number of balls of radius [1/n]. For each n, there has to be one of these, say B(x_n,[1/n]), which cannot be covered by any finite number of the sets G_i. The sequence (x_n) must have a convergent subsequence (x_nk) which converges to a limit l. Yet {G_i:i Î I} covers M, so l Î some G_i0, say.

As k ® ¥, x_nk ® l; but also [1/(n_k)] ® 0 and 1/n_k is the radius of the ball centred on x_nk. So eventually B(x_nk,[1/(n_k)]) is inside G_i0, contradictory to their choice! (More rigorously, there exists m Î w such that B(l,[2/m]) Í G_i0. Now B(l,[1/m]) contains x_nk for all k ³ k₀ say, so choose k ³ k₀ such that n_k ³ m. Then B(x_nk,[1/(n_k)]) Í B(l, [2/m]) Í G_i0.) Hence, M is compact.

2.1.5 Compactness and Uniform Continuity

Recall that a map f:(X₁,d₁) ® (X₂,d₂), where (X_i,d_i) is a metric space for each i, is uniformly continuous on X_i if given any e > 0,$d > 0 such that d₁(x,y) < d for x,y Î X₁ Þ d₂(f(x),f(y)) < e.

Ordinary continuity of f is a local property, while uniform continuity is a global property since it says something about the behaviour of f over the whole space X₁. Since compactness allows us to pass from the local to the global, the next result is not surprising:

Theorem 5 If (X,d) is a compact metric space and f:X ® R is continuous, then f is uniformly continuous on X.

Note Result holds for any metric space codomain.
Proof Let e > 0; since f is continuous, for each x Î X, $d_x > 0 such that d(x,y) < 2 d_x Þ |f(x) -f(y)| < [(e)/2]. The family {B_dx(x):x Î X } is an open cover of X and so has a finite subcover {B_dxi(x_i):i = 1,¼,n} of X. Let d = min{d_xi:i = 1,¼,n}; then, given x,y Î X such that d(x,y) < d, it follows that |f(x) -f(y)| < e

(for x Î B_dxi(x_i) for some i, whence d(x,x_i) < d_xi and so d(y,x_i) £ d(y,x) + d(x,x_i) < d+ d_xi £ 2d_xi Þ |f(y) -f(x_i)| < [(e)/2].

Thus |f(x)-f(y)| £ |f(x) -f(x_i)| + |f(x_i) - f(y)| < [(e)/2] +[(e)/2] = e).

Note Compactness is not a necessary condition on the domain for uniform continuity. For example, for any metric space (X,d), let f:X ® X be the identity map. Then f is easily seen to be uniformly continuous on X.

2.1.6 Local Compactness

Definition 4 A topological space (X,T) is locally compact iff each point of X has a compact neighbourhood.

Clearly, every compact space is locally compact. However, the converse is not true.

Examples

(i) With X infinite, the discrete space (X,D) is clearly locally compact (for each x Î X, {x} is a compact neighbourhood of x!) but not compact.
(ii) With X infinite and x Î X, (X, Á(x)) is locally compact (but not compact).
(iii) (R,Á) is locally compact (x Î R Þ [x-1,x+1] is a compact neighbourhood of x).
(iv) The set of rational numbers Q with its usual topology is not a locally compact space, for suppose otherwise; then 0 has a compact neighbourhood C in Q so we can choose e > 0 such that J = QÇ[-e, e] Í C. Now J is closed in (compact) C and is therefore compact in R. Thus, J must be closed in R-but [`J]^R = [-e,e]!

Lemma 6

(i) Local compactness is closed-hereditary.
(ii) Local compactness is preserved by continuous open maps - it is not preserved by continuous maps in general. Consider id_Q:(Q,D) ® (Q,Á_Q) which is continuous and onto; (Q, D) is locally compact while (Q, Á_Q) isn't!

Proof Exercise.

2.2 Other Covering Conditions

Definition 5 A topological space (X, T) is said to be

(i) Lindelöf iff every open cover of X has a countable subcover
(ii) countably compact iff every countable open cover of X has a finite subcover.

Thus, a space is compact precisely when it is both Lindelöf and countably compact. Further, every sequentially compact space is countably compact, although the converse is not true. Moreover, sequential compactness neither implies nor is implied by compactness.

However, for metric spaces, or more generally, metrizable spaces, the conditions compact, countably compact and sequentially compact are equivalent.
Note Second countable Þ separable; separable + metrizable Þ second countable ... and so in metrizable spaces, second countability and separability are equivalent.

2.3 Connectedness

2.3.1 Definition of Connectedness

A partition of (X, T) means a pair of disjoint, non-empty, T-open subsets whose union is X. Notice that, since these sets are complements of one another, they are both closed as well as both open. Indeed, the definition of 'partition' is not affected by replacing the term 'open' by 'closed'.

Definition 6 A connected space (X, T) is one which has no partition. (Otherwise, (X, T) is said to be disconnected.)

If Æ ¹ A Í (X, T), we call A a connected set in X whenever (A, T_A) is a connected space.

Lemma 7 (X, T) is connected iff X and Æ are the only subsets which are clopen.

Examples

(i) (X, T₀) is connected.

(ii) (X, D) cannot be connected unless |X| = 1. (Indeed the only connected subsets are the singletons!)

(iii) The Sorgenfrey line R_s is disconnected (for [x,¥) is clopen!).

(iv) The subspace Q of (R, Á) is not connected because

Q Ç[-Ö2, Ö2] closed in Q = Q Ç(-Ö2, Ö2) open in Q

is clopen and is neither universal nor empty.

(v) (X, C) is connected except when X is finite; indeed, every infinite subset of X is connected.

(vi) (X, L) is connected except when X is countable; indeed, every uncountable subset of X is connected.

(vii) In (R, Á), A Í R is connected iff A is an interval. (Thus, subspaces of connected spaces are not usually connected - examples abound in (R, Á).)

2.3.2 Characterizations of Connectedness

Lemma 8 Æ Ì A Í (X, T) is not connected iff there exist T-open sets G, H such that A Í G ÈH, A ÇG ¹ Æ, A ÇH ¹ Æ and A ÇG ÇH = Æ. (Again, we can replace 'open' by 'closed' here.)

Proof Exercise.

Note By an interval in R, we mean any subset I such that whenever a < b < c and whenever a Î I and c Î I then b Î I. It is routine to check that the only ones are (a,b), [a,b], [a,b), (a,b], [a, ¥), (a, ¥), (-¥,b), (-¥,b], (-¥,¥) = R and {a} for real a, b, a < b where appropriate.

It turns out that these are exactly the connected subsets of (R, Á):-

Lemma 9 In R, if [a,b] = F₁ ÈF₂ where F₁, F₂ are both closed and a Î F₁, b Î F₂ then F₁ ÇF₂ ¹ Æ.

Proof Exercise.

Theorem 6 Let Æ Ì I Í (R, Á). Then I is connected iff I is an interval.

Proof

$\Rightarrow$: If I is not an interval, then there exist a < b < c with a Î I, b \not Î I and c Î I. Take A = I Ç(-¥,b) and B = I Ç(b, ¥). Then A ÈB = I, A ÇB = Æ, A ¹ Æ, B ¹ Æ, A Ì I, B Ì I and A, B are both open in I i.e. A and B partition I and so I is not connected.
$\Leftarrow$: Suppose I is not connected and that I is an interval. By the `closed' version of Lemma 2.3.2, there exist closed subsets K₁, K₂ of R such that I Í K₁ ÈK₂, I ÇK₁ ¹ Æ, I ÇK₂ ¹ Æ and I ÇK₁ ÇK₂ = Æ. Select a Î I ÇK₁, b Î I ÇK₂; without loss of generality, a < b. Then [a,b] Í I so that [a,b] = ([a,b] ÇK₁) È([a,b] ÇK₂), whence by Lemma 2.3.2, Æ ¹ [a,b] ÇK₁ ÇK₂ Í I ÇK₁ ÇK₂ = Æ!

2.3.3 Connectedness and Continuous Maps

Lemma 10 Connectedness is preserved by continuous maps.

Proof Exercise.

Corollary 2 [Intermediate Value Theorem] If f:[a,b] ® R is continuous and f(a) < y < f(b), then y must be a value of f.

Proof Exercise.

Corollary 3 [Fixed point theorem for [0,1]] If f:[0,1] ® [0,1] is continuous, then it has a `fixed point' i.e. there exists some x Î [0,1] such that f(x) = x.

Proof Consider g(x) = f(x)-x. Then g:[0,1] ® R is continuous. Further, g(0) = f(0) ³ 0 and g(1) = f(1)-1 £ 0 so that 0 is intermediate between g(0) and g(1). Thus, by the Intermediate Value Theorem, there exists x Î [0,1] such that 0 = g(x) = f(x)-x i.e. such that f(x) = x.

Note Given continuous h:[a,b] ® [a,b], it follows that h has a fixed point since [a,b] @ [0,1] and `every continuous function has a fixed point' is a homeomorphic invariant.

Lemma 11 Let (X,T) be disconnected with Æ Ì Y Ì X, Y clopen. If A is any connected subset of X, then A Í Y or A Í X \Y.

Proof If A ÇY ¹ Æ ¹ A Ç(X \Y), then Æ Ì A ÇY Ì A and A ÇY is clopen in A. Thus, A is not connected! It follows that either A ÇY = Æ or A ÇX \Y = Æ i.e. either A Í X \Y or A Í Y.

Lemma 12 If the family {A_i:i Î I} of connected subsets of a space (X,T) has a non-empty intersection, then its union È_{i Î I}A_i is connected.

Proof Suppose not and that there exists a non-empty proper clopen subset Y of È_{i Î I}A_i. Then for each i Î I, either A_i Í Y or A_i Í È_{i Î I}A_i \Y. However if for some j, A_j Í Y, then A_i Í Y for each i Î I (since Ç_{i Î I}A_i ¹ Æ) which implies that È_{i Î I}A_i Í Y!

Similarly, if for some k Î I, A_k Í È_{i Î I}A_i \Y, then È_{i Î I}A_i Í È_{i Î I}A_i \Y!

Corollary 4 Given a family {C_i:i Î I} of connected subsets of a space (X,T), if B Í X is also connected and B ÇC_i ¹ Æ for all i Î I, then B È(È_{i Î I}C_i) is connected.

Proof Take A_i = B ÈC_i in Lemma 2.3.3.

Lemma 13 If A is a connected subset of a space (X,T) and A Í B Í [`A]^T, then B is a connected subset.

Proof If B is not connected, then there exists Æ Ì Y Ì B which is clopen in B . By Lemma 2.3.3, either A Í Y or A Í B \Y. Suppose A Í Y (a similar argument suffices for A Í B \Y); then [`A]<sup>T</sup> Í [`Y]^T and so B \Y = B Ç(B \Y) = [`A]<sup>T<sub>B</sub></sup> Ç(B \Y) Í [`Y]^T_B Ç(B \Y) = Y Ç(B \Y) = Æ - a contradiction!

Definition 7 Let (X, T) be a topological space with x Î X; we define the component of x, C_x, in (X, T) to be the union of all connected subsets of X which contain x i.e.

For each x Î X, it follows from Lemma 2.3.3 that C_x is the maximum connected subset of X which contains x. Also it is clear that if x, y Î X, either C_x = C_y or C_x ÇC_y = Æ (for if z Î C_x ÇC_y, then C_x ÈC_y Í C_z Í C_x ÇC_y whence C_x = C_y( = C_z)). Thus we may speak of the components of a space (X, T) (without reference to specific points of X): they partition the space into connected closed subsets (by Lemma 2.3.3) and are precisely the maximal connected subsets of X.

Examples
(i) If (X, T) is connected, (X, T) has only one component, namely X!
(ii) For any discrete space, the components are the singletons.
(iii) In Q (with its usual topology), the components are the singletons. (Thus, components need not be open.)

Definition 8 A space (X, T) is totally disconnected iff the only connected subsets of X are the singletons (equivalently, the components of (X, T) are the singletons).

Thus, by the previous examples, we see that the space Q of rationals, the space R \Q of irrationals and any discrete space are all totally disconnected. Further, the Sorgenfrey line R_s is totally disconnected.

2.3.4 Pathwise Connectedness

Definition 9 A topological space (X, T) is pathwise connected iff for any x, y Î X, there exists a continuous function f:[0,1] ® X such that f(0) = x and f(1) = y. Such a function f is called a path from x to y.

Theorem 7 Every pathwise connected space is connected.

Proof Let (X, T) be pathwise connected and let a Î X; for every x Î X, there exists a path p_x:[0,1] ® X from a to x. Then, for each x Î X, p_x([0,1]) is connected; moreover, p_a(0) = a Î Ç_{x Î X}p_x([0,1]) so that by Lemma 2.3.3, X = È_{x Î X}p_x([0,1]) is connected.

Note well The converse is false. Consider the following example, the topologist's sine curve :

(for suppose, w.l.o.g., there exists a path p:[0,1] ® X with p(0) = ([1/(p)],0) and p(1) = (0,0). Then p₁°p, being continuous, must take all values between 0 and [1/(p)], in particular [1/((2n+[1/2])p)] for each n i.e. there exists t_n Î [0,1] such that p₁°p(t_n) = [1/((2n+[1/2])p)] for each n. Thus, p(t_n) = ([1/((2n+[1/2])p)],1) ® (0,1) as n ® ¥. Now t_n Î [0,1] for all n which implies that there exists a subsequence (t_nk) in [0,1] with t_nk ® l. Then p(t_nk) ® p(l) and so p₁°p(t_nk) ® 0. Thus p(l) = (0,y) for some y, whence y = 0 (since p(l) Î X)!)

2.4 Separability

Definition 10 A topological space is said to be

(i) separable iff it has a countable dense subset.
(ii) completely separable (equivalently,second countable) iff it has a countable base.

Examples

(i) (R, Á) is separable (since [`Q] = R).
(ii) (X, C) is separable for any X.
(iii) (R, L) is not separable.

Theorem 8

(i) Complete separability implies separability.
(ii) The converse is true in metric spaces.

Proof We prove only (ii). In metric space (M,d), let D = {x₁,x₂,¼} be dense. Consider B = {B(x_i,q):i Î w,q Î Q, q > 0 }, a countable collection of open sets. One can show that B is a base for T_d ... over to you!

Theorem 9

(i) Complete separability is hereditary.
(ii) Separability is not hereditary. (Consider the `included point' topology Á(0) on R; then (R, Á(0)) is separable, since [` {0}] = R. However, R \{0} is not separable because it is discrete.)

Example
Separability does not imply complete separability since, for example, (R, Á(0)) is separable but not completely separable.(Suppose there exists a countable base B for its topology. Given x ¹ 0, {0,x } is an open neighbourhood of x and so there exists B_x Î B such that x Î B_x Í { 0,x }.Thus B_x = {0,x } i.e. B is uncountable ... contradiction!

Theorem 10 Separability is preserved by continuous maps.

Proof Exercise.

Note Complete separability is not preserved by continuous maps.

Chapter 3
Convergence

In Chapter 1, we defined limits of sequences in a topological space (X, T) so as to assimilate the metric definition. We noted, however, that not everything we knew about this idea in metric spaces is valid in topological spaces.

We will examine two main ways around this difficulty:

• develop a kind of `super-sequence' or net which does for general topology what ordinary sequences do for metric spaces.
• identify the class of topological spaces in which the old idea of sequential limit is good enough.

3.1 The Failure of Sequences

The following important results are probably familiar to us in the context of metric spaces, or at least in the setting of the real line, R.

Theorem 1 Given (X,T), A Í X, p Î X: if there exists some sequence of points of A tending to p, then p Î [`A].

Theorem 2 Given (X,T), A Í X: if A is closed, then A includes the limit of every convergent sequence of points of A.

Theorem 3 Given f:(X,T) ® (Y,T^¢): if f is continuous, then f `preserves limits of sequences' i.e. whenever x_n ® l in X, then f(x_n) ® f(l) in Y.

In each case above, it is routine to prove the statement true in a general topological space as asserted. We illustrate by proving Theorem 3.1:

Let f be continuous and x_n ® l in X. We must show that f(x_n) ® f(l). Given a neighbourhood N of f(l), there exists open G such that f(l) Î G Í N. Then l Î f^-1(G) Í f^-1(N) i.e. f^-1(N) is a neighbourhood of l and so x_n Î f^-1(N) "n ³ n₀ say. Thus f(x_n) Î N "n ³ n₀, whence f(x_n) ® f(l).

In metric spaces, the converses of these results are also true but our main point here is that in general topology, the converses are not valid.

Example
In (R, L), [`(0,1)] = R. So, for example, 5 Î [`(0,1)] and yet the only way a sequence (x_n) converges to a limit l is for x_n = l from some stage on. So no sequence in (0,1) can converge to 5 proving that the converse of Theorem 3.1 is false.

Continuing, the limit of any convergent sequence in (0,1) must belong to (0,1) for the same reason and yet (0,1) is not closed. Thus, Theorem 3.1's converse is false.

Further, id_R:(R, L) ® (R, Á) is not continuous and yet it does preserve limits of sequences.

Now this is a great nuisance! Sequences are of immense usefulness in real analysis and in metric spaces and elsewhere - and their failure to describe general topology adequately is a technical handicap. What to do?

3.2 Nets - A Kind of `Super-Sequence'

Recall that a sequence is just a function having the positive integers as domain. The set of positive integers, of course, possesses a particularly simple ordering; there is a first member, second member, third member, etc. Not all sets are naturally endowed with so simple an ordering. For example, dictionary (lexographical) ordering of words is more complex (though still relative nice as orderings go). By replacing the domain of positive integers with a set having a more complicated ordering we will:

• define a net (in analogy with sequence),
• identify an associated notion of convergence,
• show that net convergence is sufficient to characterize closure of sets,
• and that compactness can be characterized in terms of convergence of subnets.

Note that these last two items generalize the role of sequences in a metric space.

3.2.1 Definition of Net

Definition 1 A binary relation £ on a set P is said to be a pre-order iff

(i) p £ p "p Î P
(ii) p £ q and q £ r imply p £ r "p,q,r Î P.

We often refer to P as being a pre-ordered set when it is understood that £ is the pre-order in question.

If it is also true that for p,q Î P,

{\em (iii)} p £ q and q £ p imply p = q, P is said to be a partially ordered set (or poset).

Definition 2 A pre-ordered set P is said to be directed (or updirected) iff each pair of members of P has an upperbound.

(i.e. if p,q Î P, then there exists s Î P such that p £ s, q £ s.)

Definition 3 Let (P, £ ) be a poset. Then if x,y Î P with x \not £ y and y \not £ x, we write x ||y and say that x and y are incomparable;

If E Í P, then E is said to be totally unordered (or diverse) iff x,y Î E implies x = y or x ||y.

If C Í P, then C is said to be linear (or a chain or a total order) iff x,y Î C implies x < y, x = y or y < x.

(P, £ ) is said to be a lattice iff each pair of members of P has a greatest lower bound and a least upper bound.

A lattice (P, £ ) is said to be complete iff every non-empty subset Y of P has a greatest lower bound (ÙY) and a least upper bound (ÚY).

An element v of a poset (P, £ ) is said to be maximal ( minimal) iff v £ x (x £ v), x Î P Þ v = x.

Definition 4 A net in a (non-empty set) X is any function x:A ® X whose domain A is a directed set.

In imitation of the familiar notation in sequences, we usually write the net value x(a) as x_a. A typical net x:A ® X will usually appear as (x_a, a Î A) or (x_a)_{a Î A} or some such notation.

Examples of Nets

(i) N, Z, N ×N are all directed sets, where suitable pre-orders are respectively the usual magnitude ordering for N and Z, and (i,j) £ (m,n) iff i £ m and j £ n, in N ×N. Thus, for example, a sequence is an example of a net.
(ii) The real function f:R \{0} ® R given by f(x) = 3-[1/x] is a net, since its domain is a chain. Any real function is a net.
(iii) Given x Î (X, T), select in any fashion an element x_N from each neighbourhood N of x; then (x_N)_{N Î Nx} is a net in X ( since it defines a mapping from (N_x, £ ) into X). Recall that N_x is ordered by inverse set inclusion!

3.2.2 Net Convergence

Definition 5 A net (x_a)_{a Î A} in (X, T) converges to a limit l if for each neighbourhood N of l, there exists some a_N Î A such that x_a Î N for all a ³ a_N.

In such a case, we sometimes say that the net (x_a)_{a Î A}

eventuates N. Clearly, this definition incorporates the old definition of `limit of a sequence'. The limit of the net f described in (ii) above is 3. In (iii), the net described converges to x no matter how the values x_N are chosen ... prove!

3.2.3 Net Convergence and Closure

Our claim is that nets `fully describe' the structure of a topological space. Our first piece of evidence to support this is that with nets, instead of sequences, Theorems 3.1, 3.1 and 3.1 have workable converses:

Theorem 4 Given (X,T), A Í X, p Î X: p Î [`A] iff there exists a net in A converging to p.

Proof If some net of points of A converges to p, then every neighbourhood of p contains points of A (namely, values of the net) and so we get p Î [`A]. Conversely, if p is a closure point of A then, for each neighbourhood N of p, it will be possible to choose an element a_N of A that belongs also to N. The net which these choices constitute converges to p, as required.

Theorem 5 Given (X,T), A Í X, A is closed iff it contains every limit of every (convergent) net of its own points.

Proof This is really just a corollary of the preceding theorem.

Theorem 6 Given f:(X,T) ® (Y,T^¢), f is continuous iff f preserves net convergence.

Proof Exercise.

3.2.4 Nets and Compactness

Definition 6 Let (x_a)_{a Î A} be any net and let a₀ Î A. The a₀^th tail of the net is the set {x_a: a ³ a₀} = x([a₀, )). We denote it by x(a₀ ®).

Definition 7 Let (x_a)_{a Î A} and (y_b)_{b Î B} be any two nets. We call (y_b)_{b Î B} a subnet of (x_a)_{a Î A} provided that every tail of (x_a) contains a tail of (y_b) i.e. provided:

We expected a definition like `subsequence' to turn up here and we are disappointed that it has to be so complicated.

Net theory ceases to be a straightforward generalisation of sequence theory precisely when we have to take a subnet ... so we'll try to avoid this whenever possible! There is however one result certainly worth knowing:

Theorem 7 (X, T) is compact iff in X, every net has (at least one) convergent subnet.

(So, for example, (n) is a net in R with no convergent subnet.)
Proof Not required.

Corollary 1 Compactness is closed-hereditary

Proof (for if (x_a) is a net in a closed set F Í X, then it has a convergent subnet (y_b) in X. Thus there exists a subnet (z_g) of (y_b) in F which converges in X, whence its limit is in F).

Corollary 2 Compactness is preserved by continuous maps

Proof (for if X is compact and f continuous, let (y_a)_{a Î A} be a net in f(X). Then for each a Î A, y_a = f(x_a) for some x_a Î X. The net (x_a)_{a Î A} has a convergent subnet (z_b)_{b Î B}, say z_b ® l, whence f(z_b) ® f(l). Then (f(z_b))_{b Î B} is a convergent subnet of (y_a)_{a Î A}).

Example
If (x_nk) is a subsequence of a sequence (x_n), then it is a subnet of it also; because the i₀^th tail of the sequence (x_n) is

Lemma 1 If a net (x_a) converges to a limit l, then so do all its subnets.

Proof Let (y_b) be a subnet of (x_a); let N be a neighbourhood of l. Then there exists a₀ such that x_a Î N for all a ³ a₀. Further, there exists b₀ such that {y_b: b ³ b₀} Í {x_a:a ³ a₀} and so y_b Î N for all b ³ b₀.

3.3 First Countable Spaces - Where Sequences Suffice

• identify the local characteristic of topological space that makes these proofs work,
• and prove that sequences suffice to describe the topological structure of spaces with this characteristic.

3.3.1 First Countable Spaces

Definition 8 Let x Î (X, T). A countable neighbourhood base at x means: a sequence N₁, N₂, N₃, ... of particular neighbourhoods of x such that every neighbourhood of x shall contain one of the N_i's.

Note that we may assume that N₁ Ê N₂ Ê N₃ Ê ¼ because, if not, then we can work with N₁, N₁ ÇN₂, N₁ ÇN₂ ÇN₃, ...

Definition 9 We call (X, T) first-countable when every point in X has a countable neighbourhood base.

Example The classic example of a first-countable space is any metric (or metrizable) space because if x Î (M,d), then B(x,1), B(x, [1/2]), B(x, [1/3]), ... is a countable neighbourhood base at x.

Theorem 8 First-countability is hereditary and preserved by continuous open onto maps.

Proof Left to the reader.

Theorem 9

(i) Complete separability implies first countability.
(ii) Converse not always true.
(iii) Converse valid on a countable underlying set.

Proof

(i) If B is a countable base for (X, T) and p Î X, consider { B Î B: p Î B} which is a countable family of neighbourhoods of p. Moreover, they form a neighbourhood base at p.
(ii) An uncountable discrete space is first countable ( since metrizable), yet is not completely separable.
(iii) Suppose X countable and (X, T) first countable. For each x Î X, choose a countable neighbourhood base: N(x,1), N(x,2), N(x,3), .... Each is a neighbourhood of x and so contains an open neighbourhood of x: G(x,1), G(x,2), G(x,3), ....

Then B = { G(x,n) :n Î N, x Î X} is a countable family of open sets and is a base for (X, T). Thus, (X, T) is completely separable.

Example
The Arens-Fort space (see, for example, Steen and Seebach, Counterexamples in Topology is not first-countable because otherwise it would be completely separable which is false!

3.3.2 Power of Sequences in First Countable Spaces

Theorem 10 Given a first-countable space (X, T)

(i) p Î X, A Í X, then p Î [`A] iff there exists a sequence of points of A converging to p.
(ii) A Í X is closed iff A contains every limit of every convergent sequence of its own points.
(iii) f:(X,T) ® (Y,T^¢) is continuous iff it preserves limits of (convergent) sequences.

Proof

(i) Theorem 3.1 said that if there exists a sequence in A converging to some p Î X, then p Î [`A].

Conversely, if p Î [`A], then p has a countable base of neighbourhoods N<sub>1</sub> Ê N<sub>2</sub> Ê N<sub>3</sub> Ê ¼, each of which must intersect A. So choose x<sub>j</sub> Î N<sub>j</sub> ÇA for all j ³ 1. Then (x<sub>j</sub>) is a sequence in A and, given any neighbourhood H of p, H must contain one of the N<sub>j</sub>'s i.e. H Ê N<sub>j0</sub> Ê N<sub>j0 + 1</sub> Ê ¼ so that x<sub>j</sub> Î H for all j ³ j<sub>0</sub>. That is, x<sub>j</sub> ® p.
(ii) Corollary of (i).
(iii) f continuous implies that it must preserve limits of sequences (by Theorem 3.1). Conversely, if f is not continuous, there exists A Í X such that f([`A]) \not Í [`f(A)]. Thus, there exists p Î f([`A]) \[`f(A)] so p = f(x), some x Î [`A]. So there exists a sequence (x_n) in A with x_n ® x.

Yet, if f(x_n) ® f(x) ( = p), p would be the limit of a sequence in f(A) so that p Î [`f(A)] -contradiction! Thus f fails to preserve convergence of this sequence.

Chapter 4
Product Spaces

A common task in topology is to construct new topological spaces from other spaces. One way of doing this is by taking products. All are familiar with identifying the plane or 3-dimensional Euclidean space with ordered pairs or triples of numbers each of which is a member of the real line. Fewer are probably familar with realizing the torus as ordered pairs of complex numbers of modulus one. In this chaper we answer two questions:

• How do the above product constructions generalize to topological spaces?
• What topological properties are preserved by this construction?

4.1 Constructing Products

The process of constructing a product falls naturally into two stages.

• The first stage, which is entirely set-theoretic, consists in describing an element of the underlying set of the product. This task is primarily one of generalizing the notion of ordered pair or triple.
• The second stage is describing what open sets look like. This will be done by describing a subbasis for the topology. The guiding goal is to provide just enough opens sets to guarantee the continuity of certain important functions.

4.1.1 Set-Theoretic Construction

Suppose throughout that we are given a family of topological spaces {(X_i, T_i): i Î I } where I is some non-empty `labelling' or index set.

Our first task is to get a clear mental picture of what we mean by the product of the sets X_i. Look again at the finite case where I = {1,2, ¼,n}. Here, the product set

Then a typical element of X = ÕX_i will look like: (x_i)_{i Î I} or just (x_i). We will still call x_i the i^th coordinate of (x_i)_{i Î I}. ( Note that the Axiom of Choice assures us that ÕX_i is non-empty provided none of the X_i's are empty.)

4.1.2 Topologizing the Product

Definition 1 For each i Î I, the i^th projection is the map p_i:ÕX_i ® X_i which `selects the i^th coordinate' i.e. p_i((x_i)_{i Î I}) = x_i.

An open cylinder means the inverse projection of some non-empty T_i-open set i.e. p_i^-1(G_i) where i Î I, G_i ¹ Æ, G_i Î T_i.

An open box is the intersection of finitely many open cylinders Ç_{j = 1}ⁿ p_ij^-1(G_ij). The only drawable case I = {1,2} may help explain:
(Here will be, eventually, a picture!)

We use these open cylinders and boxes to generate a topology with just enough open sets to guarantee that projection maps will be continuous. Note that the open cylinders form a subbase for a certain topology T on X = ÕX_i and therefore the open boxes form a base for T; T is called the [Tychonoff]product topology and (X, T) is the product of the given family of spaces. We write (X,T) = Õ{ (X_i, T_i): i Î I } = Õ_{i Î I}(X_i, T_i) or even T = Õ_{i Î I}T_i.

Notice that if Ç_{j = 1}ⁿp_ij^-1(G_ij) is any open box, then without loss of generality we can assume i₁, i₂, ... i_n all different because if there were repetitions like

It is routine to check that if T_n is the usual topology on Rⁿ, and T the usual topology on R, then

Lemma 1 In a product space (X, T), N is a neighbourhood of p Î X iff there exists some open box B such that p Î B Í N.

Lemma 2 For each i Î I,

(i) p_i is continuous
(ii) p_i is an open mapping.

Proof

(i) Immediate.
(ii) Given open G Í X, then G is a union of basic open sets {B_k:k Î K} in X, whence p_i(G) is a union of open subsets {B_kⁱ:k Î K} of X_i and is therefore open. (The notation here is intended to convey that B_kⁱ is the 'component along the i-th coordinate axis' of the open box B_k.)

Theorem 1 A map into a product space is continuous iff its composite with each projection is continuous.

Proof Since the projections are continuous, so must be their composites with any continuous map. To establish the converse, first show that if S is a subbase for the codomain (target) of a mapping f, then f will be continuous provided that the preimage of every member of S is open; now use the fact that the open cylinders constitute a subbase for the product topology.

Worked example Show that (X, T) ×(Y, S) is homeomorphic to (Y, S) ×(X, T).
Solution
Define f:X ×Y ® Y ×X and g:Y ×X ® X ×Y by f(x,y) = (y,x) and g(y,x) = (x,y). Clearly these are one-one, onto and mutually inverse. It will suffice to show that both are continuous.

p₁°f = p₂^¢; p₂°f = p₁^¢. Now p_i^¢ is continuous for i = 1,2 and so f is continuous! Similarly, g is continuous.

Worked example Show that the product of infinitely many copies of (N, D) is not locally compact.
Solution
We claim that no point has a compact neighbourhood. Suppose otherwise; then there exists p Î X, C Í X and G Í X with C compact, G open and p Î G Í C. Pick an open box B such that p Î B Í G Í C. B looks like Ç_{j = 1}ⁿp_ij^-1(G_ij). Choose i_n+1 Î I \{i₁,i₂, ¼,i_n}; then p_in+1(C) is compact (since compactness is preserved by continuous maps).

Thus, p_in+1 Î p_in+1(B) = X_in+1 Í p_in+1(C) Í X_in+1 = (N, D). Thus, p_in+1(C) = (N, D) ¼ which is not compact!

4.2 Products and Topological Properties

4.2.1 Products and Connectedness

Theorem 2 Any product of connected spaces must be connected.

Proof is left to the reader.

4.2.2 Products and Compactness

Theorem 3 [Tychonoff's theorem] Any product of compact spaces is compact i.e. compactness is productive.

Proof It suffices to prove that any covering of X by open cylinders has a finite subcover. Suppose not and let C be a family of open cylinders which covers X but for which no finite subcover exists. For each i Î I, consider

To prove the above without Alexander's Subbase Theorem is very difficult in general, but it is fairly simple in the special case where I is finite. Several further results show that various topological properties are 'finitely productive' in this sense.

Theorem 4 If (X₁, T₁), (X₂, T₂), ..., (X_n, T_n) are finitely many sequentially compact spaces, then their product is sequentially compact.

Proof
Take any sequence (x_n) Î X. The sequence ( p₁(x_n))_{n ³ 1} in sequentially compact X₁ has a convergent subsequence p₁(x_nk) ® l₁ Î X₁. The sequence (p₂(x_nk))_{k ³ 1} in sequentially compact X₂ has a convergent subsequence (p₂(x_nkj))_{j ³ 1} ® l₂ Î X₂ and p₁(x_nkj) ® l₁ also.

Do this n times! We get a subsequence (y_p)_{p ³ 1} of the original sequence such that p_i(y_p) ® l_i for i = 1,2, ¼, n. It's easy to check that y_p ® (l₁, l₂, ¼, l_n) so that X is sequentially compact, as required.

Lemma 3 `The product of subspaces is a subspace of the product.'

Proof
Let (X, T) = Õ_{i Î I}(X_i, T_i); let Æ Ì Y_i Í X_i for each i Î I. There appear to be two different ways to topologise ÕY_i:

{\em either} (i) give it the subspace topology induced by ÕT_i
{\em or} (ii) give it the product of all the individual subspace topologies (T_i)_Yi.

The point is that these topologies coincide-if G_i0^* is open in (T_i0)_Yi0 where i₀ Î I i.e. G_i0^* = Y_i0 ÇG_i0 for some G_i0 Î T_i0, a typical subbasic open set for (ii) is

Theorem 5 Local compactness is finitely productive.

Proof
Given x = (x₁,x₂,¼,x_n) Î (X, T) = Õ_{i = 1}ⁿ(X_i,T_i), we must show that x has a compact neighbourhood. Now, for all i = 1, ¼, n, x_i has a compact neighbourhood C_i in (X_i,T_i) so we choose T_i-open set G_i such that x_i Î G_i Í C_i. Then

Lemma 4 [`(ÕY<sub>i</sub>)]<sup>T</sup> = Õ[`(Y_i)]^T_i ( in notation of previous lemma).

Proof Do it yourself! (`The closure of a product is a product of the closures.')

4.2.3 Products and Separability

Theorem 6 Separability is finitely productive.

Proof
For 1 £ i £ n, choose countable D_i Í X_i where [`(D<sub>i</sub>)]<sup>T<sub>i</sub></sup> = X<sub>i</sub>. Consider D = D<sub>1</sub> ×D<sub>2</sub> ×¼×D<sub>n</sub> = Õ<sub>1</sub><sup>n</sup>D<sub>i</sub>, again countable. Then [`D] = [`(ÕD<sub>i</sub>)]<sup>T</sup> = Õ[`(D_i)]^T_i = ÕX_i = X.

Notice that the converses of all such theorems are easily true. For example,

Theorem 7 If (X, T) = Õ_{i Î I}(X_i, T_i) is

(i) compact
(ii) sequentially compact
(iii) locally compact
(iv) connected
(v) separable
(vi) completely separable

then so is every `factor space' (X_i, T_i).

Proof For each i Î I, the projection mapping p_i: X ® X_i is continuous, open and onto. Thus, by previous results, the result follows.

Chapter 5
Separation Axioms

We have observed instances of topological statements which, although true for all metric (and metrizable) spaces, fail for some other topological spaces. Frequently, the cause of failure can be traced to there being `not enough open sets' (in senses to be made precise). For instance, in any metric space, compact subsets are always closed; but not in every topological space, for the proof ultimately depends on the observation

`given x ¹ y, it is possible to find disjoint open sets G and H with x Î G and y Î H'

which is true in a metric space (e.g. put G = B(x,e), H = B(y,e) where e = [1/2] d(x,y)) but fails in, for example, a trivial space (X, T₀).

What we do now is to see how `demanding certain minimum levels-of-supply of open sets' gradually eliminates the more pathological topologies, leaving us with those which behave like metric spaces to a greater or lesser extent.

5.1 T₁ Spaces

Definition 1 A topological space (X,T) is T₁ if, for each x in X, {x} is closed.

Comment 1

(i) Every metrizable space is T₁
(ii) (X,T₀) isn't T₁ unless |X| = 1

Theorem 1

(i) T₁ is hereditary
(ii) T₁ is productive
(iii) T₁ Þ every finite set is closed. More precisely, (X, T) is T₁ iff T Ê C, i.e. C is the weakest of all the T₁ topologies that can be defined on X.

Proof is left to the reader.

The respects in which T₁-spaces are `nicer' than others are mostly concerned with `cluster point of a set' (an idea we have avoided!). We show the equivalence, in T₁ spaces, of the two forms of its definition used in analysis.

Theorem 2 Given a T₁ space (X,T), p Î X and A Í X, the following are equivalent:

(i) Every neighbourhood of p contains infinitely many points of A
(ii) Every neighbourhood of p contains at least one point of A different from p.

Proof Obviously, (i) Þ (ii); conversely, suppose (i) fails; so there exists a neighbourhood N of p such that N ÇA is finite. Consider H = [X\(NÇA)] È{ p }; it is cofinite and is thus an (open) neighbourhood of p. Hence N ÇH is a neighbourhood of p which contains no points of A, except possibly p itself. Thus, (ii) fails also.

Hence, (i) Û (ii).

5.2 T₂ (Hausdorff) Spaces

Definition 2 A topological space (X,T) is T₂ (or Hausdorff) iff given x ¹ y in X, $ disjoint neighbourhoods of x and y.

Comment 2

(i) Every metrizable space is T₂
(ii) T₂ Þ T₁ (i.e. any T₂ space is T₁, for if x, y Î T₂ X and y Î [`{x}], then every neighbourhood of y contains x, whence x = y.)
(iii) (X,C), with X infinite, cannot be T₂

Theorem 3

(i) T₂ is hereditary
(ii) T₂ is productive.

Proof

{i} The proof is left to the reader.
{ii} Let (X, T) = Õ_{i Î I}(X_i, T_i) be any product of T₂ spaces. Let x = (x_i)_{i Î I} and y = (y_i)_{i Î I} be distinct elements of X. Then there exists i₀ Î I such that x_i0 ¹ y_i0 in X_i0. Choose disjoint open sets G, H in (X_i0, T_i0) so that x_i0 Î G, y_i0 Î H. Then x Î p_i0^-1(G) Î T, y Î p_i0^-1(H) Î T and since G ÇH = Æ, p_i0^-1(G)Çp_i0^-1(H) = Æ. Hence result.

Theorem 4 In a T₂-space (X, T), if C is a compact set and x \not Î C, then there exist T-open sets G and H so that x Î G, C Í H and G ÇH = Æ.

Proof A valuable exercise: separate each point of C from x using disjoint open sets, note that the open neighbourhoods of the various elements of C, thus obtained, make up an open covering of C, reduce it to a finite subcover by appealing to compactness ...

Corollary 1 In a T₂-space, any compact set is closed.

Corollary 2 In a T₂-space, if C and K are non-empty compact and disjoint, then there exist open G, H such that C Í G, K Í H and G ÇH = Æ.

A basic formal distinction between algebra and topology is that although the inverse of a one-one, onto group homomorphism [etc!] is automatically a homomorphism again, the inverse of a one-one, onto continuous map can fail to be continuous. It is a consequence of Corollary 5.2 that, amongst compact T₂ spaces, this cannot happen.

Theorem 5 Let f:(X₁,T₁) ® (X₂,T₂) be one-one, onto and continuous, where X₁ is compact and X₂ is T₂. Then f is a homeomorphism.

Proof It suffices to prove that f is closed. Given closed K Í X₁, then K is compact whence f(K) is compact and so f(K) is closed. Thus f is a closed map.

Theorem 6 (X,T) is T₂ iff no net in X has more than one limit.

Proof

(i) $\Rightarrow$ (ii): Let x ¹ y in X; by hypothesis, there exist disjoint neighbourhoods U of x, V of y. Since a net cannot eventually belong to each of two disjoint sets, it is clear that no net in X can converge to both x and y.
(ii) $\Rightarrow$ (i): Suppose that (X, T) is not Hausdorff and that x ¹ y are points in X for which every neighbourhood of x intersects every neighbourhood of y. Let N_x (N_y) be the neighbourhood systems at x (y) respectively. Then both N_x and N_y are directed by reverse inclusion. We order the Cartesian product N_x ×N_y by agreeing that

(Ux,Uy) ³ (Vx,Vy) Û Ux Í Vx and Uy Í Vy.

Evidently, this order is directed. For each (U_x, U_y) Î N_x ×N_y, U_x ÇU_y ¹ Æ and hence we may select a point z_(Ux,Uy) Î U_x ÇU_y. If W_x is any neighbourhood of x, W_y any neighbourhood of y and (U_x,U_y) ³ (W_x,W_y), then

z(Ux,Uy) Î Ux ÇUy Í Wx ÇWy.

That is, the net { z_(Ux,Uy),(U_x,U_y) Î N_x ×N_y} eventually belongs to both W_x and W_y and consequently converges to both x and y!

Corollary 3 Let f:(X₁, T₁) ® (X₂, T₂), g:(X₁, T₁) ® (X₂, T₂) be continuous where X₂ is T₂. Then their `agreement set' is closed i.e. A = {x:f(x) = g(x)} is closed.

5.3 T₃ Spaces

Definition 3 A space (X, T) is called T₃ or regular provided :-

(i) it is T₁, and
(ii) given x \not Î closed F, there exist disjoint open sets G and H so that x Î G, F Í H.

Comment 3

(i) Every metrizable space is T₃; for it is certainly T₁ and given x \not Î closed F, we have x Î open X \F so there exists e > 0 so that x Î B(x, e) Í X \F. Put G = B(x, [(e)/2]) and H = { y :d(x,y) > [(e)/2] }; the result now follows.
(ii) Obviously T₃ Þ T₂.
(iii) One can devise examples of T₂ spaces which are not T₃.
(iv) It's fairly routine to check that T₃ is productive and hereditary.
(v) Warning: Some books take T₃ to mean Definition 5.3(ii) alone, and regular to mean Definition 5.3(i) and (ii); others do exactly the opposite!

5.4 T_3[1/2] Spaces

Definition 4 A space (X, T) is T_3[1/2] or completely regular or Tychonoff iff

(i) it is T₁, and
(ii) given x Î X, closed non-empty F Í X such that x \not Î F, there exists continuous f:X ® [0,1] such that f(F) = {0} and f(x) = 1.

Comment 4

(i) Every metrizable space is T_3[1/2]
(ii) Every T_3[1/2] space is T₃ ( such a space is certainly T₁ and given x \not Î closed F, choose f as in the definition; define G = f^-1([0, [1/3])), H = f^-1(([2/3],1]) and observe that T₃ follows.)
(iii) Examples are known of T₃ spaces which fail to be Tychonoff
(iv) T_3[1/2] is productive and hereditary.

5.5 T₄ Spaces

Definition 5 A space (X, T) is T₄ or normal if

(i) it is T₁, and
(ii) given disjoint non-empty closed subsets A, B of X, there exist disjoint open sets G, H such that A Í G, B Í H.

Theorem 7 Every metrizable space (X, T) is T₄.

Proof Certainly, X is T₁; choose a metric d on X such that T is T_d. The distance of a point p from a non-empty set A can be defined thus:

G = {x:d(x, A) < d(x,B) }
H = {x:d(x,B) < d(x,A)}.

Clearly, G ÇH = Æ. Also, each is open (if x Î G and e = [1/2]{d(x,B) - d(x,A)}, then B(x,e) Í G, by the triangle inequality.) Now, if d(p, A) = 0, then for all n Î N, there exists x_n Î A such that d(p, x_n) < [1/n]. So d(p,x_n)® 0 i.e. x_n ® p, whence p Î [`A]. Thus for each x Î A, x \not Î B = [`B] so that d(x, B) > 0 = d(x,A) i.e. x Î G. Hence A Í G. Similarly B Í H.

It's true that T₄ Þ T_3[1/2] but not very obvious. First note that if G₀, G₁ are open in a T₄ space with [`(G<sub>0</sub>)] Í G<sub>1</sub>, then there exists open G<sub>[1/2]</sub> with [`(G₀)] Í G_[1/2] and [`(G<sub>[1/2]</sub>)] Í G<sub>1</sub> (because the given [`(G₀)] and X \G₁ are disjoint closed sets so that there exist disjoint open sets G_[1/2], H such that [`(G₀)] Í G_[1/2], X\G₁ Í H i.e. G₁ Ê (closed) X\H Ê G_[1/2]).

Lemma 1 [Urysohn's Lemma] Let F₁, F₂ be disjoint non-empty closed subsets of a T₄ space; then there exists a continuous function f:X ® [0,1] such that f(F₁) = {0}, f(F₂) = {1}.

Proof Given disjoint closed F₁ and F₂, choose disjoint open G₀ and H₀ so that F₁ Í G₀, F₂ Í H₀. Define G₁ = X \F₂ (open). Since G₀ Í (closed) X \H₀ Í X \F₂ = G₁, we have [`(G₀)] Í G₁.

By the previous remark, we can now construct:

(i) G_[1/2] Î T: [`(G<sub>0</sub>)] Í G<sub>[1/2]</sub>, [`(G_[1/2])] Í G₁.
(ii) G_[1/4], G_[3/4] Î T: [`(G<sub>0</sub>)] Í G<sub>[1/4]</sub>, [`(G_[1/4])] Í G_[1/2], [`(G<sub>[1/2]</sub>)] Í G<sub>[3/4]</sub>, [`(G_[3/4])] Í G₁.
(iii) ... and so on!

Thus we get an indexed family of open sets

Observe that the index set is dense in [0,1]: if s < t in [0,1], there exists some [m/(2ⁿ)] such that s < [m/(2ⁿ)] < t. Define

Well, f(x) < a iff there exists some r = [m/(2ⁿ)] such that f(x) < r < a. It follows that f^-1([0, a)) = È_{r < a}G_r, a union of open sets.

Again, f(x) > a iff there exist r₁, r₂ such that a < r₁ < r₂ < f(x), implying that x \not Î G_r2 whence x \not Î [`(G<sub>r1</sub>)]. It follows that f<sup>-1</sup>((a,1]) = È<sub>r1 > a</sub>(X \[`(G_r1)]), which is again open.

Corollary 4 Every T₄ space is T_3[1/2].

Proof Immediate from Lemma 5.5. (Note that there exist spaces which are T_3[1/2] but not T₄.)

Theorem 8 Any compact T₂ space is T₄.

Proof Use Corollary 5.2 to Theorem 5.2.

Note Unlike the previous axioms, T₄ is neither hereditary nor productive. The global view of the hierarchy can now be filled in as an exercise from data supplied above:-

Metrizable	Hereditary?	Productive?
T4
T3[1/2]
T3
T2
T1

The following is presented as an indication of how close we are to having `come full circle'.

Theorem 9 Any completely separable T₄ space is metrizable!

Sketch Proof
Choose a countable base; list as { (G_n,H_n): n ³ 1} those pairs of elements of the base for which [`(G<sub>n</sub>)] Í H<sub>n</sub>. For each n, use Lemma 5.5 to get continuous f<sub>n</sub>:X ® [0,1] such that f<sub>n</sub>([`(G_n)]) = {0}, f_n(X \H_n) = {1}. Define