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.

<img src="http://mathonline.wdfiles.com/local--files/boundedness-of-compact-sets-in-a-metric-space/Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" alt="Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" class="image" />

• 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. $\blacksquare$

Licensing

Content obtained and/or adapted from:

• Compact Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license