Topology: An Inquiry-Based Approach

Let $Q_0^{\prime }$ be one of these sub-cubes. The collection $\{O_{\alpha }\cap Q_0^{\prime }{\}}_{\alpha \in I}$ is an open cover of $Q_0^{\prime }$ in the subspace topology. If each of these open covers has a finite sub-cover, then we can take the union of all of the $O_{\alpha }$s over all of the finite sub-covers to obtain a finite sub-cover of $\{O_{\alpha }{\}}_{\alpha \in I}$ for $Q_0\text{.}$ Since our cover $\{O_{\alpha }{\}}_{\alpha \in I}$ for $Q_0$ has no finite sub-cover, we conclude that there is one sub-cube, $Q_1\text{,}$ for which the open cover $\{O_{\alpha }\cap Q_1{\}}_{\alpha \in I}$ has no finite sub-cover. Now we repeat the process and partition $Q_1$ into $2^n$ uniform sub-cubes of side length $\frac{M}{2}=\frac{2M}{2^2}\text{.}$ The same argument we just made tells us that there is a sub-cube $Q_2$ of $Q_1$ for which the open cover $\{O_{\alpha }\cap Q_2{\}}_{\alpha \in I}$ has no finite sub-cover (an illustration for the $n=2$ case is shown at middle in Figure 17.11). We proceed inductively to obtain an infinite nested sequence

of cubes such that for each $k\in \mathbb{Z}\text{,}$ the lengths of the sides of cube $Q_k$ are $\frac{M}{2^{k-1}}=\frac{2M}{2^k}$ and the open cover $\{O_{\alpha }\cap Q_k{\}}_{\alpha \in I}$ of $Q_k$ has no finite sub-cover. Now we show that $\bigcap _{k=1}^{\mathrm{\infty }}{Q}_k\ne \mathrm{\varnothing }\text{.}$

For $i\in {\mathbb{Z}}^{+}\text{,}$ let $Q_i=[a_{i,1},b_{i,1}]\times [a_{i,2},b_{i,2}]\times \cdots [a_{i,n},b_{i,n}]\text{.}$ That is, think of the point $(a_{i,1},a_{i,2},\dots ,a_{i,n})$ as a lower corner of the cube and the point $(b_{i,1},b_{i,2},\dots ,b_{i,n})$ as an upper corner of the $n$-cube $Q_i$ (a labeling for $n=2$ and $i$ from 1 to 3 is shown at right in Figure 17.11). Let $q=(sup\{a_{i,1}\},sup\{a_{i,2}\},\dots ,sup\{a_{i,n}\})\text{.}$ We will show that $q\in \bigcap _{k=1}^{\mathrm{\infty }}{Q}_k\text{.}$ Fix $r\in {\mathbb{Z}}^{+}\text{.}$ We need to demonstrate that

For each $s$ between 1 and $n$ we have

because $sup\{a_{i,s}\}$ is an upper bound for all of the $a_{i,s}\text{.}$ The fact that our cubes are nested means that

for every $i$ and $s\text{.}$ Since $sup\{a_{i,s}\}$ is the least upper bound of all of the $a_{i,s}\text{,}$ property (17.2) shows that $sup\{a_{i,s}\}\le b_{i,s}$ for every $i\text{.}$ Thus, $sup\{a_{i,s}\}\le b_{r,s}$ and so $a_{r,s}\le sup\{a_{i,s}\}\le b_{r,s}\text{.}$ This shows that $q\in Q_k$ for every $k\text{.}$ Consequently, $q\in \bigcap _{k=1}^{\mathrm{\infty }}{Q}_k$ and $\bigcap _{k=1}^{\mathrm{\infty }}{Q}_k$ is not empty. (The fact that the side lengths of our cubes are converging to 0 implies that $\bigcap _{k=1}^{\mathrm{\infty }}{Q}_k=\{q\}\text{,}$ but we only need to know that $\bigcap _{k=1}^{\mathrm{\infty }}{Q}_k$ is not empty for our proof.)

Since $\{O_{\alpha }{\}}_{\alpha \in I}$ is a cover for $Q_0\text{,}$ there must exist an ${\alpha }_q\in I$ such that $q\in O_{{\alpha }_q}\text{.}$ The set $O_{{\alpha }_q}$ is open, so there exists ${ϵ}_q>0$ such that $B(q,{ϵ}_q)\subseteq O_{{\alpha }_q}\text{.}$ The maximum distance between points in $Q_k$ is the distance between the corner points $(a_{k,1},a_{k,2},\dots ,a_{k,n})$ and $(b_{k,1},b_{k,2},\dots ,b_{k,n})\text{,}$ where each length $b_{k,s}-a_{k,s}$ is $\frac{M}{2^{k-1}}\text{.}$ The distance formula tells us that this maximum distance between points in $Q_k$ is

For the converse, assume that $A$ is a compact subset of ${\mathbb{R}}^n\text{.}$ We must show that $A$ is closed and bounded. Now $({\mathbb{R}}^n,d_E)$ is a metric space, and so Hausdorff. Theorem 17.6 then shows that $A$ is closed. To conclude our proof, we need to demonstrate that $A$ is bounded. For each $k>0\text{,}$ let

That is, $O_k$ is the open $k$-cube in ${\mathbb{R}}^n\text{.}$ Next let

for each $k\text{.}$ Since $\bigcup _{k>0}{O}_k={\mathbb{R}}^n\text{,}$ it follows that $\{U_k{\}}_{k>0}$ is an open cover of $A\text{.}$ The fact that $A$ is compact means that there is a finite collection $U_{k_1}\text{,}$ $U_{k_2}\text{,}$ $\dots \text{,}$ $U_{k_m}$ of sets in $\{U_k{\}}_{k>0}$ that cover $A\text{.}$ Let $K=max\{k_i\mid 1\le i\le m\}\text{.}$ Then $U_{k_i}\subseteq U_K$ for each $i\text{,}$ and so $A\subseteq U_K\subset Q_K^m\text{.}$ Thus, $A$ is bounded. This completes the proof that if $A$ is compact in ${\mathbb{R}}^n\text{,}$ then $A$ is closed and bounded.