Compactness

Proposition 422.1(Proposition).

1. Every closed subspace of a compact space is compact.

2. Every compact subspace of a Hausdorff space is closed.

3. The image of a compact space under a continuous map is compact.

4. Let $f:X\to Y$ be a bijective continuous function. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

Proof.

$(2)$ Proof proceed in the same way as that of Prp 347.8.3; show that every point not in the subspace has a neighborhood in the complement.

$(4)$ If $A\subseteq X$ is closed in $X$, then $A$ is compact. Therefore, $f(A)$ is compact. Since $Y$ is Hausdorff, $f(A)$ is closed in $Y$.□

Lemma 422.2(Tube lemma).

Consider the product space $X\times Y$, where $Y$ is compact. If $N$ is an open set of $X\times Y$ containing the slice $x_0\times Y$ of $X\times Y$, then $N$ contains some tube $W\times Y$ about $x_0\times Y$, where $W$ is a neighborhood of $x_0$ in $X$.

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Lem 2 is used in proving that a finite product of compact spaces is compact.

Compactness generalizes the Cantor intersection property; see Finite intersection property. Also, the uniform continuity theorem.

The order topology provides us with examples of compact spaces analogous to the metric case.

Proposition 422.3(Proposition).

Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.

We also have the following generalization of the extreme value theorem:

Proposition 422.4(Extreme value theorem).

Let $f:X\to Y$ be continuous, where $Y$ is an ordered set in the order topology. If $X$ is compact, then there exist points $c$ and $d$ in $X$ such that $f(c)⩽f(x)⩽f(d)$ for every $x\in X$.

For metric spaces, we get a refinement of compactness: the Lebesgue number lemma.

Remark 422.5(Remark).

If a metric space $(X,d)$ satisfies the conclusion of the Lebesgue number lemma, then
any continuous map from $X$ to a metric space is uniformly continuous.

Theorem 422.6(Theorem).

Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

Proof.

First, show that given any nonempty open set $U$ of $X$ and any point $x$ of $X$, there exists a nonempty open set $V$ contained in $U$ such that $x\notin \overline{V}$; this used the Hausdorff hypothesis. Then, show that given $f:{\mathbb{Z}}_{+}\to X$, $f$ is not surjective by constructing a descending chain of closed intervals ${\overline{V}}_1\supseteq {\overline{V}}_2\supseteq \dots$ such that ${\overline{V}}_n$ does not contain $x_n$. Using the finite intersection property, $\bigcap {\overline{V}}_n$ is nonempty, and the element contained in this intersection cannot be in the image of $f$.□

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Limit point compactness

Proposition 422.7(Proposition).

Compactness implies limit point compactness, but not conversely.

Proof.

Prove the contrapositive - if $A$ has no limit point, then $A$ must be finite. If $A$ has no limit point, it is closed and all points are isolated.□

Example 422.8(Example).

Let $Y$ consist of tow points and have the indiscrete topology. Then the space $X={\mathbb{Z}}_{+}\times Y$ is limit point compact, but not compact.

For metric spaces, compactness, limit point compactness, and sequential compactness are equivalent.

Remark 422.9(Remark).

For a Hausdorff space, we have
compact ⇒ countably compact ⇔ limit point compact
sequentially compact ⇒ countably compact ⇔ limit point compact

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Local compactness

Definition 422.10(Local compactness).

A space $X$ is called locally compact at $x$ if there is a compact subspace $C$ of $X$ that contains a neighborhood $V$ of $x$: that is, $x\in V\subseteq C\subseteq X$. A space is locally compact if it is locally compact at $x$ for all $x\in X$.

Theorem 422.11(Theorem).

Let $X$ be a space. Then $X$ is locally compact Hausdorff iff there exists a space $Y$ satisfying the following conditions:

1. $X$ is a subspace of $Y$.
2. The set $Y\setminus X$ consists of a single point.
3. $Y$ is a compact Hausdorff space.

If $Y$ and $Y^{\prime }$ are two spaces satisfying these conditions, then there is a homeomorphism of $Y$ with $Y^{\prime }$ that equals the identity map on $X$.

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Definition 422.12(Definition).

If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y\setminus X$ equals a single point, then $Y$ is called the one-point compactification of $X$; the use of ‘the’ is justified by Thm 11.

If we assume $X$ is Hausdorff, local compactness permits the kind of formulation we’d expect for a ‘local’ property:

Proposition 422.13(Proposition).

Let $X$ be a Hausdorff space. Then $X$ is locally compact iff for each $x\in X$ and each neighborhood $U\ni x$, there is a neighborhood $V\ni x$ such that $\overline{V}$ is compact and is contained in $U$: that is, $x\in V\subseteq \overline{V}\subseteq U$.

Proposition 422.14(Proposition).

Let $X$ be locally compact Hausdorff. Let $A\subseteq X$. If $A$ is closed in $X$ or open in $X$, then $A$ is locally compact.

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Corollary 422.15(Corollary).

A space $X$ is homeomorphic to an open subspace of a compact Hausdorff space iff $X$ is locally compact Hausdorff; follows from Thm 11 and Prp 14.

Proposition 422.16(Munkres (2000) §32 Exr 3).

Every locally compact Hausdorff space is regular.

Proof.

X is locally compact Hausdorff, so by Thm 11 it has a one point compactification which is compact Hausdorff hence normal by Prp 347.8.3. Then $X$ is a subspace of a normal space and hence regular by Prp 347.7.3. In fact, this shows that $X$ is completely regular since Prp 421.4.□

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References

Munkres, J. R. (2000). Topology (2nd ed). Prentice Hall, Inc.