Compactness in Metric Spaces

Compact Sets in a Metric Space

If $(M,d)$ is a metric space and $S\subseteq M$ then a cover or covering of $S$ is a collection of subsets ${\mathcal {F}}$ in $M$ such that:

${\begin{aligned}\quad S\subseteq \bigcup _{A\in {\mathcal {F}}}A\end{aligned}}$

Furthermore, we said that an open cover (or open covering) is simply a cover that contains only open sets.

We also said that a subset ${\mathcal {S}}\subseteq {\mathcal {F}}$ is a subcover/subcovering (or open subcover/subcovering if ${\mathcal {F}}$ is an open covering) if ${\mathcal {S}}$ is also a cover of $S$, that is:

${\begin{aligned}\quad S\subseteq \bigcup _{A\in {\mathcal {S}}}A\quad {\text{ where }}{\mathcal {S}}\subseteq {\mathcal {F}}\end{aligned}}$

We can now define the concept of a compact set using the definitions above.

Definition: Let $(M,d)$ be a metric space. The subset $S\subseteq M$ is said to be Compact if every open covering ${\mathcal {F}}$ of $S$ has a finite subcovering of $S$.

In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. So, let's look at an example of a subset of a metric space that is not compact.

Consider the metric space $(\mathbb {R} ,d)$ where $d$ is the Euclidean metric and consider the set $S=[0,1)\subseteq \mathbb {R}$. We claim that this set is not compact. To show that $S$ is not compact, we need to find an open covering ${\mathcal {F}}$ of $S$ that does not have a finite subcovering. Consider the following open covering:

${\begin{aligned}\quad {\mathcal {F}}=\left\{\left((0,1-{\frac {1}{n}}\right):n\in \mathbb {Z} ,n\geq 1\right\}=\left\{\left(0,{\frac {1}{2}},\right),\left(0,{\frac {2}{3}}\right),\left(0,{\frac {3}{4}}\right),...\right\}\end{aligned}}$

Clearly ${\mathcal {F}}$ is an infinite subcovering of $(0,1)$ and furthermore:

${\begin{aligned}\quad (0,1)\subseteq \bigcup _{k=2}^{\infty }\left(0,1-{\frac {1}{k}}\right)\end{aligned}}$

Let ${\mathcal {F}}^{*}$ be a finite subset of ${\mathcal {F}}$ containing $p$ elements. Then:

${\begin{aligned}\quad {\mathcal {F}}^{*}=\left\{\left(0,1-{\frac {1}{n_{1}}}\right),\left(0,1-{\frac {1}{n_{2}}}\right),...,\left(0,1-{\frac {1}{n_{p}}}\right)\right\}\end{aligned}}$

Let $n^{*}=\max\{n_{1},n_{2},...,n_{p}\}$. Then due to the nesting of the open covering ${\mathcal {F}}$, we see that:

${\begin{aligned}\quad \bigcup _{k=1}^{p}\left(0,1-{\frac {1}{n_{p}}}\right)=\left(0,1-{\frac {1}{n^{*}}}\right)\end{aligned}}$

But for $(0,1)\subseteq \left(0,1-{\frac {1}{n^{*}}}\right)$ we need $1\leq 1-{\frac {1}{n^{*}}}$. But $n^{*}\in \mathbb {N}$, so $n^{*}>0$ and ${\frac {1}{n^{*}}}>0$, so $1-{\frac {1}{n^{*}}}<1$. Therefore any finite subset ${\mathcal {F}}^{*}$ of ${\mathcal {F}}$ cannot cover $S=(0,1)$. Hence, $(0,1)$ is not compact.

Boundedness of Compact Sets in a Metric Space

Recall that if $(M,d)$ is a metric space then a subset $S\subseteq M$ is said to be compact in $M$ if for every open covering of $S$ there exists a finite subcovering of $S$.

We will now look at a rather important theorem which will tell us that if $S$ is a compact subset of $M$ then we can further deduce that $S$ is also a bounded subset.

Theorem 1: If $(M,d)$ be a metric space and $S\subseteq M$ is a compact subset of $M$ then $S$ is bounded.

• Proof: For a fixed $x_{0}\in S$ and for $r>0$, consider the ball centered at $x_{0}$ with radius $r$, i.e., $B(x_{0},r)$. Let ${\mathcal {F}}$ denote the collection of balls centered at $x_{0}$ with varying radii $r>0$:

${\begin{aligned}\quad {\mathcal {F}}=\{B(x_{0},r):r>0\}\end{aligned}}$

• It should not be hard to see that ${\mathcal {F}}$ is an open covering of $S$, since for all $s\in S$ we have that $d(x_{0},s)=r_{s}>0$, so $s\in B(x_{0},r_{s})\in {\mathcal {F}}$.

• Now since $S$ is compact and since ${\mathcal {F}}$ is an open covering of $S$, there exists a finite open subcovering subset ${\mathcal {F}}^{*}\subset {\mathcal {F}}$ that covers $S$. Since ${\mathcal {F}}^{*}$ is finite, we have that:

${\begin{aligned}\quad {\mathcal {F}}^{*}=\{B(x_{0},r_{1}),B(x_{0},r_{2}),...,B(x_{0},r_{p})\}\end{aligned}}$

• And by definition ${\mathcal {F}}^{*}$ covers $S$ so:

${\begin{aligned}\quad S\subseteq \bigcup _{k=1}^{p}B(x_{0},r_{k})\end{aligned}}$

• Each of the open balls in the open subcovering ${\mathcal {F}}^{*}$ is centered at $x_{0}$ with $r_{1},r_{2},...,r_{p}>0$. Since the set $\{r_{1},r_{2},...,r_{p}\}$ is a finite set, there exists a maximum value. Let:

${\begin{aligned}\quad r_{\mathrm {max} }=\max\{r_{1},r_{2},...,r_{p}\}\end{aligned}}$

• Then for all $k\in \{1,2,...,p\}$ we have that $B(x_{0},r_{k})\subseteq B(x_{0},r_{\mathrm {max} })$ and therefore:

${\begin{aligned}\quad S\subseteq \bigcup _{k=1}^{p}B(x_{0},r_{k})=B(x_{0},r_{\mathrm {max} })\end{aligned}}$

• Hence $S$ is bounded. $\blacksquare$

Basic Theorems Regarding Compact Sets in a Metric Space

Recall that if $(M,d)$ is a metric space then a set $S\subseteq M$ is said to be compact in $M$ if for every open covering of $S$ there exists a finite subcovering of $S$.

We will now look at some theorems regarding compact sets in a metric space.

Theorem 1: Let $(M,d)$ be a metric space and let $S,T\subseteq M$. Then if $S$ is closed and $T$ is compact in $M$ then $S\cap T$ is compact in $M$.

• Proof: Let $S$ be closed and let $T$ be compact in $M$. Notice that:

${\begin{aligned}\quad S\cap T\subseteq T\end{aligned}}$

• Furthermore, $S\cap T$ is closed. This is because $S$ is given as closed, and since $T$ is compact we know that $T$ is closed (and bounded). So the finite intersection $S\cap T$ is closed. But any closed subset of a compact set is also compact, so $S\cap T$ is compact in $M$. $\blacksquare$

Theorem 2: Let $(M,d)$ be a metric space and let $S_{1},S_{2},...,S_{n}\subseteq M$ be a finite collection of compact sets in $M$. Then $\displaystyle {\bigcup _{i=1}^{n}S_{i}}$ is also compact in $M$.

• Proof: Let $S_{1},S_{2},...,S_{n}\subseteq M$ be a finite collection of compact sets in $M$. Consider the union $S=\bigcup _{i=1}^{n}S_{i}$ and let ${\mathcal {F}}$ be any open covering of $S$, that is:

${\begin{aligned}\quad S\subseteq \bigcup _{A\in {\mathcal {F}}}A\end{aligned}}$

• Now since $S_{i}\subseteq S$ for all $i\in \{1,2,...,n\}$ we see that ${\mathcal {F}}$ is also an open covering of $S$ and so there exists a finite subcollection ${\mathcal {F}}_{i}\subseteq {\mathcal {F}}$ that also covers $S_{i}$, i.e.:

${\begin{aligned}\quad S_{i}\subseteq \bigcup _{A\in {\mathcal {F}}_{i}}A\end{aligned}}$

• Let ${\mathcal {F}}^{*}=\bigcup _{i=1}^{n}{\mathcal {F}}_{i}$. Then ${\mathcal {F}}^{*}$ is finite since it is equal to a finite union of finite sets. Furthermore:

${\begin{aligned}\quad S=\bigcup _{i=1}^{n}S_{i}\subseteq \bigcup _{i=1}^{n}\left(\bigcup _{A\in {\mathcal {F}}_{i}}A\right)=\bigcup _{A\in {\mathcal {F}}^{*}}A\end{aligned}}$

• So ${\mathcal {F}}^{*}\subseteq {\mathcal {F}}$ is a finite open subcovering of $S$. So for all open coverings ${\mathcal {F}}$ of $S$ there exists a finite open subcovering of $S$, so $\displaystyle {S=\bigcup _{i=1}^{n}S_{i}}$ is compact in $M$. $\blacksquare$

Theorem 3: Let $(M,d)$ be a metric space and let ${\mathcal {C}}$ be an arbitrary collection of compact sets in $M$. Then $\displaystyle {\bigcap _{C\in {\mathcal {C}}}C}$ is also compact in $M$.

• Proof: Let ${\mathcal {C}}$ be an arbitrary collection of compact sets in $M$. Notice that for all $C\in {\mathcal {C}}$ that:

${\begin{aligned}\quad \bigcap _{C\in {\mathcal {C}}}C\subseteq C\end{aligned}}$

• Furthermore, since each $C\in {\mathcal {C}}$ is compact, then each $C$ is closed (and bounded). An arbitrary intersection of closed sets is closed, and so $\displaystyle {\bigcap _{C\in {\mathcal {C}}}}$ is a closed subset of the compact set $C$. Therefore by the theorem referenced earlier, $\displaystyle {\bigcap _{C\in {\mathcal {C}}}C}$ is compact in $M$. $\blacksquare$

Licensing

Content obtained and/or adapted from:

• Compact Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license
• Boundedness of Compact Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license
• Basic Theorems Regarding Compact Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license