Topology: An Inquiry-Based Approach

Let \(Q'_0\) be one of these sub-cubes. The collection \(\{O_{\alpha} \cap Q'_0\}_{\alpha \in I}\) is an open cover of \(Q'_0\) 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 \(k \in \Z\text{,}\) 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 \(\bigcap_{k=1}^{\infty} Q_k \neq \emptyset\text{.}\)

For \(i \in \Z^+\text{,}\) 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 \((a_{i,1}, a_{i,2}, \ldots, a_{i,n})\) as a lower corner of the cube and the point \((b_{i,1}, b_{i,2}, \ldots, b_{i,n})\) 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 \(q = (\sup\{a_{i,1}\}, \sup\{a_{i,2}\}, \ldots, \sup\{a_{i,n}\})\text{.}\) We will show that \(q \in \bigcap_{k=1}^{\infty} Q_k\text{.}\) Fix \(r \in \Z^+\text{.}\) 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 \(\sup\{a_{i,s}\} \leq b_{i,s}\) for every \(i\text{.}\) Thus, \(\sup\{a_{i,s}\} \leq b_{r,s}\) and so \(a_{r,s} \leq \sup\{a_{i,s}\} \leq b_{r,s}\text{.}\) This shows that \(q \in Q_k\) for every \(k\text{.}\) Consequently, \(q \in \bigcap_{k=1}^{\infty} Q_k\) and \(\bigcap_{k=1}^{\infty} Q_k\) is not empty. (The fact that the side lengths of our cubes are converging to 0 implies that \(\bigcap_{k=1}^{\infty} Q_k = \{q\}\text{,}\) but we only need to know that \(\bigcap_{k=1}^{\infty} Q_k\) 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 \(\epsilon_q \gt 0\) such that \(B(q, \epsilon_q) \subseteq O_{\alpha_q}\text{.}\) The maximum distance between points in \(Q_k\) is the distance between the corner points \((a_{k,1}, a_{k,2}, \ldots, a_{k,n})\) and \((b_{k,1}, b_{k,2}, \ldots, b_{k,n})\text{,}\) 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 \(\R^n\text{.}\) We must show that \(A\) is closed and bounded. Now \((\R^n, d_E)\) 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 \(k \gt 0\text{,}\) let

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

for each \(k\text{.}\) Since \(\bigcup_{k \gt 0} O_k = \R^n\text{,}\) it follows that \(\{U_k\}_{k \gt 0}\) is an open cover of \(A\text{.}\) The fact that \(A\) is compact means that there is a finite collection \(U_{k_1}\text{,}\) \(U_{k_2}\text{,}\) \(\ldots\text{,}\) \(U_{k_m}\) of sets in \(\{U_k\}_{k \gt 0}\) that cover \(A\text{.}\) Let \(K = \max\{k_i \mid 1 \leq i \leq m\}\text{.}\) Then \(U_{k_i} \subseteq U_K\) for each \(i\text{,}\) and so \(A \subseteq U_K \subset Q^m_K\text{.}\) Thus, \(A\) is bounded. This completes the proof that if \(A\) is compact in \(\R^n\text{,}\) then \(A\) is closed and bounded.