CAL1_L7

Finite intersection property

Definition 1(Definition).

A collection $F$ of sets in $X$ is said to have the finite intersection property if any finite subcollection of $F$ has a nonempty intersection.

If $T$ is a collection of open subsets of a metric space $X$, then the collection $F$ of complements of sets in $T$ is a collection of closed sets. Moreover, $T$ is a cover of $X$ if and only if $F$ has empty intersection. Now, $X$ is compact if $T$ being an open cover implies a finite subcover of $T$ exists. Taking the contrapositive, we get that $X$ is compact if no finite subcover of $T$ implies $T$ is not an open cover.

Theorem 2(Theorem).

A metric space $X$ is compact if and only if every collection $F$ of closed subsets of $X$ with the finite intersection property has nonempty intersection.

Characterization of compact sets

Recall definitions of open cover compactness (referred to as just compactness), sequential compactness, and limit point compactness. In Analysis 1, we showed that sequential and limit point compactness are equivalent. We stated but didn’t show that this equivalence extends to open cover compactness. We will prove this. Also recall that we showed that all compact sets are closed and bounded (with the converse being true for ${\mathbb{R}}^n$). We will introduce a stronger characterization, which says that a set is compact if and only if it is complete and totally bounded.

Definition 3(Definition).

A metric space $X$ is said to be totally bounded if for each $ϵ>0$, $X$ can be covered by a finite number of open balls of radius $ϵ$. A subset $E$ of $X$ is said to be totally bounded if $E$ is bounded as a metric space.

A totally bounded metric space is always bounded. However, being bounded does not imply being totally bounded. For example, consider $\mathbb{R}$ with the discrete metric. $\mathbb{R}$ is bounded under the discrete metric, but is not totally bounded since a finite covering of $ϵ$-balls does not exist for $ϵ<1$. But, under any metric induced by a norm, a subset of ${\mathbb{R}}^n$ is bounded iff it is totally bounded.

Theorem 4(Theorem).

A subset of ${\mathbb{R}}^n$ is bounded iff it is totally bounded under any metric induced by a norm.

Proof
Since we have shown that all norms on ${\mathbb{R}}^n$ are equivalent, it will suffice to prove the theorem for the metric induced by the euclidean norm.
Let $E$ be a bounded subset of ${\mathbb{R}}^n$. Since $E$ is bounded, we may take $a$ large enough such that $E\subset [-a,a{]}^n$. Choose $k$ such that $\sqrt{n}a/k<ϵ$. Let $P_k$ be a partition of $[-a,a]$ for which each partition has length less than $\frac{1}{k}$. Then $P_k^n$ induces a partition of $[-a,a{]}^n$ into $n$-cuboids of diameter less than $\sqrt{n}a/k$. Consider the finite collection of balls of radius $ϵ$ with centers $(x,y)$ where $x$ and $y$ are the partition points of $P_k$. Then this finite collection of balls of radius $ϵ$ covers $[-a,a{]}^n$, and hence covers $E$.

Theorem 5(Characterization of compactness for a metric space).

For a metric space $X$, the following assertions are equivalent:

• $X$ is complete and totally bounded
• $X$ is compact
• $X$ is sequentially compact.

Complete and totally bounded $⟹$ compact

We argue by contradiction. Let $\mathcal{O}=\{{\mathcal{O}}_{\lambda }{\}}_{\lambda \in \Lambda }$ be an open cover of $X$ with no finite subcover. Since $X$ is totally bounded, there exists a cover of $X$ which consists of open balls of radius $ϵ=\frac{1}{2}$. At least one of these open balls must not have an open subcover in $\mathcal{O}$. Select one such ball and label its closure $F_1$. Now take a cover of $F_1$ consisting of open balls of radius $ϵ=\frac{1}{4}$. The intersection of at least one such ball with $F_1$ must not have an open subcover in $\mathcal{O}$. Select one such ball and label the closure of its intersection with $F_1$ as $F_2$. Then $F_1$ and $F_2$ are closed, $F_2\subseteq F_1$, and $\text{diam}(F_1)<1$, $\text{diam}(F_2)<\frac{1}{2}$. Continuing in this way we obtain a contracting sequence of nonempty closed sets $(F_n)$ with the property that each $F_n$ does not have a finite subcover in $\mathcal{O}$. Since $X$ is complete, it has the Cantor intersection property, which implies ${\bigcap }_{n=1}^{\infty }{F}_n=\{x\}$, where $x\in X$.

Now, $x\in {\mathcal{O}}_{\lambda }$ for some $\lambda \in \Lambda$. Since $(F_n)$ is a contracting sequence and ${\mathcal{O}}_{\lambda }$ is open, $F_n\subseteq {\mathcal{O}}_{\lambda }$ for some $n$. This contradicts the choice of $F_n$ as being a set that does not have a finite subcover in $\mathcal{O}$.

Compact $⟹$ sequentially compact

Let $(x_n)$ be a sequence in $X$. For each index $n$, let $F_n$ be the closure of $\{x_k|k\ge n\}$. Note that the collection $\{F_n{\}}_{n=1}^{\infty }$ has the finite intersection property. Since $X$ is compact, it follows that ${\bigcap }_{n=1}^{\infty }{F}_n\ne \varnothing$. Let $p\in {\bigcap }_{n=1}^{\infty }{F}_n$. Now, either $p$ is a limit point of $(x_n)$, or $p\in F_n$ for all $n$. In both cases, we can extract a subsequence of $(x_n)$ which converges to $p$. Therefore, $X$ is sequentially compact.

Sequentially compact $⟹$ complete and totally bounded

Sequentially compact $⟹$ complete is obvious.
Assume $X$ is not totally bounded. Then for some $ϵ>0$, we cannot cover $X$ with a finite number of $ϵ$-balls. Pick $x_1\in X$, $x_2\in X\setminus B_{ϵ}(x_1)$, $x_3\in X\setminus B_{ϵ}(x_1)\setminus B_{ϵ}(x_2)$, and so on. The sequence $(x_n)$ does not have a convergent subsequence since the distance between any two of its points is $ϵ$ or greater.