Download LECtURE 7: SEPtEmBER 17 Closed sets and compact sets. Last

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Lecture 7: September 17
Closed sets and compact sets. Last time, we defined compactness in terms of
open coverings: a topological space X is compact if, for every open covering U ,
there are finitely many open sets U1 , . . . , Un ∈ U such that X = U1 ∪ · · · ∪ Un .
To prove that a given space is not compact is usually not difficult: we just have to
find one open covering that violates the condition in the definition. To prove that
a given space is compact can be quite difficult: we have to verify the condition for
all possible open coverings. (This task becomes a little easier when the topology
on X is given by a basis, because we only need to consider coverings by basic
open subsets.) Since compactness is a strong requirement, compact spaces have a
lot of wonderful properties. The following theorem about continuous functions on
compact spaces that shows this effect at work.
Theorem 7.1. Let f : X → Y be a continous function. If X is compact and Y is
Hausdorff, then the image f (X) is closed.
Recall from last time that the image of a compact space under a continuous
function is again compact; this was an easy consequence of the definitions. Theorem 7.1 requires a little bit more work to prove. Let me first note the following
surprising corollary.
Corollary 7.2. Let f : X → Y be a bijective continous function. If X is compact
and Y is Hausdorff, then f is a homeomorphism.
Proof. Let us see how this follows from Theorem 7.1. We have to prove is that
the inverse function f −1 : Y → X is continous; this is equivalent to saying that the
preimage of every closed set A ⊆ X is closed in Y . The preimage of A under f −1 is
exactly f (A), and so we need to argue that f (A) is closed. As a closed subset of a
compact space, A is compact (Proposition 6.13); therefore its image f (A) is again
compact by Proposition 6.14. Since Y is Hausdorff, Theorem 7.1 implies that f (A)
is closed.
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Now let us start proving the theorem. By Proposition 6.14, the image f (X) is
a compact subspace of the Hausdorff space Y . The following proposition explains
why f (X) must be closed.
Proposition 7.3. Let X be a Hausdorff topological space. If Y ⊆ X is a compact
subspace, then Y is closed in X.
Proof. We shall argue that X \ Y is a union of open sets, and hence open. Fix a
point x ∈ X \ Y . For any y ∈ Y , we can find disjoint open sets U (y) and V (y) with
x ∈ U and y ∈ V ; this is because X is Hausdorff. Therefore
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Y ⊆
V (y),
y∈C
and because Y is compact, there are finitely many points y1 , . . . , yn ∈ Y with
Y ⊆ V (y1 ) ∪ · · · ∪ V (yn ). But then the open set
U (y1 ) ∩ · · · ∩ U (yn )
is disjoint from V (y1 ) ∪ · · · ∪ V (yn ), hence contained in X \ Y ; because it is a
neighborhood of x, and because x ∈ X \ Y was arbitrary, X \ Y must be open. �
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Note that this result is false when X is not Hausdorff; a simple example is given
by the Sierpiński space. More generally, let X be any finite topological space; then
all compact subsets of X are closed if and only if the topology on X is discrete if
and only if X is Hausdorff.
Note. The proof of Proposition 7.3 shows that we can separate points and compact
subspaces in a Hausdorff space by open sets. More precisely, given a compact
subspace Y ⊆ X and a point x �∈ Y , there are disjoint open sets U and V with
x ∈ U and Y ⊆ V – in the notation used during the proof,
U = U (y1 ) ∩ · · · ∩ U (yn )
and
V = V (y1 ) ∪ · · · ∪ V (yn ).
If X is both compact and Hausdorff, so that every closed subspace is compact, we
can even separate points and arbitrary closed subspaces by open sets. We shall
investigate results of this type in more detail later.
Examples of compact spaces. Let us now look at a few topological spaces that
are compact. The first example is closed intervals in R. You probably already know
that closed intervals are “compact” in the analysis sense – every sequence has a
convergent subsequence – but we need to do some work to prove that they are also
compact in the topology sense.
Theorem 7.4. The closed interval [0, 1] is compact.
Proof. Let U be an open covering; we have to show that the interval can be covered
by finitely many open sets in U . The idea of the proof is very simple: 0 belongs
to some open set U1 ∈ U , which contains some interval of the form [0, a); then a
belongs to some open set U2 ∈ U , which again contains a maximal interval [a, b);
and so on. Of course, the situation could be like in one of Zeno’s paradoxes, with the
points a, b, . . . converging to some limit inside the interval. To avoid this problem,
we have to take a slightly less direct approach. Let us introduce the set
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S = x ∈ [0, 1] � [0, x] can be covered by finitely open sets in U .
We want to show that 1 ∈ S, because this will mean that [0, 1] can be covered
by finitely many sets in U . Using the least upper bound property of R, define
s = sup S. Now the argument proceeds in three steps:
Step 1. Every point t with 0 ≤ t < s has to belong to S. Suppose that we had
t �∈ S. Then every point x ∈ S has to satisfy x < t: the reason is that [0, x] can
be covered by finitely open sets in U , and if t ≤ x, then the same open sets would
cover [0, t]. This means that t is an upper bound for S, contradicting our choice of
s. Consequently, we have [0, s) ⊆ S.
Step 2. We show s ∈ S. Since U is an open covering, there is an open set U ∈ U
with s ∈ U . If s = 0, this shows that s ∈ S. If s > 0, then U contains an interval of
the form (a, s] with 0 ≤ a < s. By Step 1, a ∈ S, and so [0, a] is covered by finitely
many open sets in U ; throwing in U , we get a finite open covering of [0, s].
Step 3. We prove that s = 1. Suppose that s < 1. We already know that [0, s]
is covered by finitely many open sets in U . Their union is a neighborhood of s,
and therefore contains an interval of the form [s, b) with s < b ≤ 1. Because U is
an open covering, b ∈ U for some U ∈ U ; but now we have finitely many open sets
covering [0, b], contradicting our choice of s.
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In fact, every closed interval of the form [a, b] is compact: this follows either by
adapting the above proof, or by noting that any (nontrivial) closed interval in R is
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homeomorphic to [0, 1]. In the proof, we only used the ordering of R and the least
upper bound property; for that reason, the theorem is true is any linearly ordered
set with the least upper bound property.
The next example is Rn and its subsets. If you are taking analysis, you may
already know that a subspace of Rn is compact if and only if it is closed and
bounded (in the Euclidean metric). Here bounded means that it is contained in a
ball BR (0) of some radius R. I want to explain why this result is also true with
our definition of compactness. The following general result about products will be
useful for the proof.
Proposition 7.5. If X and Y are compact, then X × Y is compact.
Proof. Recall that the product topology has a basis consisting of all open sets of
the form U × V , with U open in X and V open in Y . Let U be an open covering
of X × Y ; as I explained earlier, it suffices to consider the case that U consists of
basic open sets.
Step 1. Fix a point x ∈ X, and consider the vertical slice p−1
1 (x) = {x} × Y . It
is homeomorphic to Y , and therefore compact; one way of seeing this is to consider
the continuous function i : Y → X ×Y , i(y) = (x, y), and to apply Proposition 6.14.
Since it is covered by the union of all the open sets in U , we can find finitely many
open sets U1 × V1 , . . . , Un × Vn ∈ U such that p−1
1 (x) ⊆ U1 × V1 ∪ · · · ∪ Un × Vn .
Now define U (x) = U1 ∩ · · · ∩ Un ; then the entire set U (x) × Y is covered by finitely
many open sets in U .
Step 2. The collection of open sets U (x) from Step 1 is an open covering of X.
By compactness, there are finitely many points x1 , . . . , xm ∈ X with
X = U (x1 ) ∪ · · · ∪ U (xm ).
This means that
X × Y = U (x1 ) × Y ∪ · · · ∪ U (xm ) × Y,
and because each these m sets is covered by finitely many sets in U , the same is
true for the product X × Y .
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More generally, any finite product of compact spaces is again compact; this can
easily be proved by induction on the number of factors. We will see later in the
semester that arbitrary products of compact spaces are compact (in the product
topology); this is an extremely powerful result, but the proof is not that easy.
Theorem 7.6. A subset A ⊆ Rn is compact if and only if it is closed and bounded.
Proof. Let us first check that every compact subset A ⊆ Rn is both closed and
bounded. As Rn is Hausdorff, Proposition 6.13 implies that A is closed. To prove
boundedness, observe that
∞
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A⊆
Bn (0)
n=1
is an open covering; because A is compact, we must have A ⊆ Bn (0) for some n.
The really interesting part is the converse. Suppose that A ⊆ Rn is closed and
bounded. Choose some R > 0 with the property that A ⊆ BR (0); then also
A ⊆ [−R, R]n .
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We know from Theorem 7.4 that the closed interval [−R, R] is compact; the space
on the right-hand side is therefore compact by Proposition 7.5. Being a closed
subset of a compact space, A is therefore compact as well.
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Next time, we will look at some other results about compact spaces, and at one-point
compactifications.