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# Compactness

### From Intelligent Perception

If you are looking for "compactness" as a measure of roundness of objects, then go there... |

A well known fact from Calculus is the following:

**Theorem.** Every bounded sequence has a convergent subsequence.

The proof may go like this. For any $\epsilon >0$, the elements of any sequence can be covered by a collection, possibly infinite, of $\epsilon$-intervals. Since the sequence is bounded, one can choose a finite sub-collection that still covers the sequence. Then at least on of the intervals will contain infinitely many elements of the sequence. Then you go on to prove convergence.

The crucial step in the proof is the ability to choose a finite *subcover*. This motivates the following definitions.

**Definition.** For any topological space $X$ (or a subset $X$ of some other topological space), a collection of open sets $\alpha$
is called an *open cover* if $\cup \alpha = X$ (or $X \subset \cup \alpha$).

**Definition.** A topological space $X$ is called *compact* if every open cover contains a finite subcover.

Note: The nature of the definition tells us that compactness is a topological invariant.

Let's take a finite open interval $I=(0,1)$ in ${\bf R}$:.

Suppose the cover consists of all open $\epsilon$-intervals: $$\alpha = \{(a,b) \cap I \colon b-a= \epsilon \}.$$

Then finding a finite subcover is easy: $$\alpha ' = \{(0,\epsilon),(1/2 \epsilon,(1+1/2)\epsilon), (\epsilon,2\epsilon),... \}.$$

There will be $[2/ \epsilon]+1$ intervals.

However, the following cover of $(0,1)$ does not have a finite subcover: $$\alpha = \{(a_n,b_n) \colon n=1,2,3,\ldots \},$$ if we choose $$a_n \rightarrow 1, b_n \rightarrow 1,a_n = 0,a_n < b_n.$$

Same applies to $[0,1)$, and similarly to $(0,1]$.

Thus *open and half-open intervals aren't compact*!

But closed intervals are:

**Theorem.** A closed bounded subset of a Euclidean space is compact.

This isn't true for infinite dimensional spaces. To see why, consider the sequence: $$(1,0,0,0, \ldots), (0,1,0,0, \ldots), (0,0,1,0, \ldots ), \ldots$$

Also, *unbounded subsets in the Euclidean space aren't compact.* For example, ${\bf R}$ can't even be covered by a finite collection of finite intervals.

**Exercise.** Show that the image of a compact space under a continuous function is compact.

It follows that compactness is a topological property.

**Theorem.** A closed subset of a compact space is compact.

**Proof.** Suppose $X$ is compact and $A \subset X$ is closed. Suppose $\alpha$ is an open cover of $A$. The idea is to first construct from $\alpha$ an open cover of the whole $X$. That's easy:
$$\beta = \alpha \cup \{X \setminus A \}.$$
The last set is open because $A$ is closed. Since $X$ is compact $\beta$ has a finite subcover, $β'$. Now we need to construct from $β'$ a cover for $A$. That's easy:
$$\alpha' = \beta' \setminus \{X \setminus A \}.$$

Finish the proof (**exercise**). $\blacksquare$

A partial converse is the following.

**Theorem.** Suppose $X$ is Hausdorff. If a subset $A$ of $X$ is compact, then it is closed.

Compactness makes it easier to prove topological equivalence.

**Theorem.** Suppose $X$ is compact and $Y$ is Hausdorff. If $f \colon X \rightarrow Y$ is continuous, one-to-one, and onto, then $f$ is a homeomorphism.

**Proof.** Suppose $A$ is a closed subset of $X$ which is compact. Then $A$ is compact by a theorem above. By the previous theorem the image $f(A)$ of $A$ is a compact subset of $Y$. Since $Y$ is Hausdorff, this implies that $f(A)$ is closed in $Y$. Thus, the image of any closed set under $f$ is closed. Hence the preimage of any closed set under the inverse function $f^{-1}$ of $f$ is closed. So, $f^{-1}$ is continuous. $\blacksquare$