Proof
(related to Proposition: Closed n-Dimensional Cuboids Are Compact)
- Let \(a_\nu,b_\nu\in\mathbb R\) be real numbers with \(a_\nu \le b_\nu\) for \(\nu=1,\ldots,n\).
- Consider the closed $n$-dimensional cuboid \(Q:=\{(x_1,\ldots,x_n)\in\mathbb R^n:\quad a_\nu \le x_\nu \le b_\nu\}\).
- It has to be shown that \(Q\) is a compact subset of the \(n\)-dimensional metric space of real numbers \(\mathbb R^n\).
- Take any open cover of \((U_i)_{i\in I}\) of \(Q\).
- Assume, there is no finite subcover \(U_{i_k}\) of \(Q\).
* Set \(Q_0:=Q\), and $I_{0,\nu}=[a_\nu,b_\nu]$, where $I_{0,\nu}$ are closed real intervals for \(\nu=1,\ldots,n\). Note that $Q_0=I_{0,1}\times\ldots\times I_{0,n}$.
* Bisect $I_{0,\nu}$ by setting $$I_{1,\nu}^L=\left[a_\nu,\frac {b_\nu-a_\nu}2\right],\quad I_{1,\nu}^R=\left[\frac {b_\nu-a_\nu}2, b_\nu,\right],$$ such that $I_{0,\nu}=I_{1,\nu}^L\cup I_{1,\nu}^R$, for \(\nu=1,\ldots,n\).
* From the bisected intervals, we can construct $2^n$ new cuboids $$Q_1^{(s_1,s_2,\ldots,s_n)}=I_{1,1}^{s_1}\times\ldots\times I_{1,n}^{s_n},\quad\quad ( * )$$ such that $s_\nu$ assumes the values \(L\) for $I_{1,\nu}^L$ or $R$ for $I_{1,\nu}^R$, and \(\nu=1,\ldots,n\).
* Because, by assumption, \(Q_0\) has no finite subcover, there is at least one cuboid out of the $2^n$ cuboids in $ ( * )$, which has no finite subcover. Let this cuboid be denoted by \(Q_1\).
* Please note that $$\operatorname{diam}(Q_1)=\frac 12\operatorname{diam}(Q_0),$$ i.e. the diameter of \(Q_1\) is half the diameter of \(Q_0\).
* We can repeat this construction by finding even smaller cuboids, \(Q_2,Q_3,\ldots\) such that for \(m\ge 1\) they have the following two properties:
* $Q_m$ has no finite subcover,
* $\operatorname{diam}(Q_m)=\frac 12\operatorname{diam}(Q_{m-1})=\frac 1{2^{m}}\operatorname{diam}(Q_{0})$, and
* $Q_{m-1}\supset Q_m$.
* According to the nested closed subset theorem, there is a point \(a\) with \(a\in Q_{m}\) for all \(m\in\mathbb N\).
* Since \((U_i)_{i\in I}\) is an open cover of of \(Q\), there is an index \(i_a\in I\) such that \(a\in U_{i_a}\).
* Because $U_{i_a}$ is open, it is a neighborhood of \(a\).
* Therefore there is an \(\epsilon > 0\) such that the open ball \(B(a,\epsilon)\) is contained in $U_{i_a}$.
* Take \(m\) big enough, such that $\operatorname{diam}Q_m < \epsilon$.
* Because $a\in Q_m$, we have the inclusion chain
$$Q_m\subset B(a,\epsilon)\subset U_{i_a}.$$
* This contradicts that $Q_m$ has no finite subcover.
* This contradicts, that $Q$ has no finite subcover.
- Thus, $Q$ has a finite subcover for any given open cover of \((U_i)_{i\in I}\) of \(Q\).
- Thus, \(Q\) is a compact subset of the \(n\)-dimensional metric space of real numbers \(\mathbb R^n\).
∎
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References
Bibliography
- Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984
