# Proof

• 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 $$\mathbb R^n$$.

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984