
# Chapter 3 - Topological Spaces

## 3.10 Compactness

 Main: Index Previous: 3.9 Basic Concepts Next: 3.11 Compactness in Metric Spaces

### Results

Heine-Borel Theorem

Any cover of a closed interval $[a, b]$ by a system of open interval (or, open sets) has a finite subcover.

Example
In the following, $C$ is a cover of $X$, since it's union contains $X$. $U$ is a subcover of $C$, since $U\subset C$. $$X=\{1,2,3\}$$ $$C=\{\{1\},\{2\},\{1,3\}\}$$ $$U=\{\{2\},\{1,3\}\}$$

Definition 1

A topological space $T$ is said to be compact if every open cover of $T$ has a finite subcover. A compact Hausdorff space is called a compactum.

Definition 2

A system of subsets $\{A_\alpha\}$ of a set $T$ is said to be centered if every finite intersection $\bigcap_{k=1}^n A_k$ is nonempty.

Theorem 1

A topological space $T$ is compact if and only if it has the following property:

$\Delta$) Every centered system of closed subsets of $T$ has a nonempty intersection.

Theorem 2

Every closed subset $F$ of a compact topological space $T$ is itself compact.

Corollary

Every closed subset of a compactum is itself a compactum.

Theorem 3

Let $K$ be a compactum and $T$ any Hausdorff space containing $K$. Then $K$ is closed in $T$.

Theorem 4

Every compactum $K$ is a normal space.

Theorem 5

Let $X$ be a compact space and $f$ a continuous mapping of $X$ onto a topological space $Y$. Then $Y = f(X)$ is itself compact.

Theorem 6

A one-to-one continuous mapping of a compactum $X$ onto a compactum $Y$ is necessarily a homeomorphism.

Theorem 7

If $T$ is a compact space, then any infinite subset of $T$ has at least one limit point.

Definition 3

A topological space $T$ is said to be countably compact if every infinite subset of $T$ has at least one limit point (in $T$).

Theorem 8

Each of the following two conditions are necessary and sufficient for a topological space $T$ to be countably compact.
1) Every countable open cover of $T$ has a finite subcover.
2) Every countable centered system of closed subsets of $T$ has a nonempty intersection.

Important Distinction
A compact topological space are those where an arbitrary open cover have a finite subcover, and the countably compact spaces where a countable open cover has a finite subcover. The following Theorem shows the special case where the first implies the second.

Theorem 9

The concepts of compactness and countable compactness coincide for a topological space $T$ with a countable base.

Definition 4

A subset $M$ of a topological space $T$ is said to be relatively compact (in $T$) if its closure $\overline{M}$ in $T$ is compact.

Definition 5

A subset $M$ of a topological space $T$ is said to be relatively countably compact (in $T$) if every infinite subset $A\subset M$ has at least one limit point in $T$ (which may not belong to $T$).

### Problem 1

Skipped... (Uninteresting counterexample)

### Problem 2

A topological space $T$ is said to be locally compact if every point $x\in T$ has at least one relatively compact neighborhood. Show that a compact space is automatically locally compact, but not conversely. Prove that every closed subspace of a locally compact subspace is locally compact.

Proof.
Suppose $T$ is a compact space and let $x$ be some arbitrary point in $T$. Then select some open neighborhood containing $x$, $U\ni x$. The closure $[U]$ is a closed subset of a compact space. By Theorem 2, $[U]$ will be a compact space, so $U$ is a relatively compact neighborhood. This shows that $T$ is locally compact.

Now a counterexample to show that the converse is not true; a space can be locally compact; i.e. every point has a neighborhood whose closure is compact. But that the space is not compact; an open cover does NOT have a finite subcover. An example of this is simply $\R$, which is locally compact, but not compact. Every $x\in\R$ is contained in an open set $(a, b)$ where the closure $[a, b]$ is compact, but there is no finite subcover of $\R$.

Suppose $K$ is a locally compact subspace of $T$, and let $F$ be a closed subspace of $K$. Let $x$ be a random point $x\in F$. Then $x\in K$ and since $K$ is locally compact, there is at least one neighborhood $U_x\ni x$ such that $[U_x]$ is compact with respect to $K$. Let $[V_x] = [U_x]\cap F$, then $[V_x]\subset [U_x]$ is a closed subset of a compact space $[U_x]$ and by Theorem 2, $[V_x]$ is compact. But then $V_x$ is a relatively compact neighborhood of $x$ so $F$ is locally compact.

### Problem 3

A point $x$ is said to be a complete limit point of a subset $A$ of a topological space, if given any neighborhood $U\ni x$, the sets $A$ and $A\cap U$ have the power (i.e. cardinal number). Prove that every infinite subset of a compact space $T$ has at least one complete limit point.

Proof.
Suppose $T$ is a compact space. Then any open covering $\{U_\alpha\}$ has a finite subcover, i.e. $$T \subset \bigcup_{k=1}^N U_k.$$ Now let $A\subset T$ be an infinite subset. Since $$A\subset T \subset \bigcup_{k=1}^N U_k,$$ then there is at least one open set $U_m$, $1\leq m\leq N$ that contains an infinite amount of points. (We can also assume that $U_m$ is the largest set). Take some $x\in U_m$, which is a neighborhood containing an infinite amount of points. Then $A\cap U_m$ has the same power, and $x$ is a complete limit point of $A$.