open set in ${ℝ}^n$ contains an open rectangle

Theorem Suppose ${ℝ}^n$ isequipped with the usual topology induced by the open balls of theEuclidean metric.Then, if $U$ is a non-empty open set in ${ℝ}^n$, thereexist real numbers $a_i,b_i$ for $i=1,\mathrm{\dots },n$ such that and $[a_1,b_1]\times \mathrm{\cdots }\times [a_n,b_n]$ is a subset of $U$.

Proof.Since $U$ is non-empty, there exists some point $x$in $U$. Further, since $U$ is a topological space, $x$ is contained insome open set. Since the topology has a basis consisting ofopen balls, there exists a $y\in U$ and $\epsilon >0$ such that $x$is contained in the open ball $B(y,\epsilon )$.Let us now set $a_i=y_i-\frac{\epsilon }{2\sqrt{n}}$ and$b_i=y_i+\frac{\epsilon }{2\sqrt{n}}$for all $i=1,\mathrm{\dots },n$.Then $D=[a_1,b_1]\times \mathrm{\cdots }\times [a_n,b_n]$ can beparametrized as

$$D=\{y+({\lambda }_1,\mathrm{\dots },{\lambda }_n)\frac{\epsilon }{2\sqrt{n}}\mid {\lambda }_i\in [-1,1]\text{for all}i=1,\mathrm{\dots },n\}.$$

For an arbitrary point in $D$, we have

$|y+({\lambda }_1,\mathrm{\dots },{\lambda }_n){\displaystyle \frac{\epsilon }{2\sqrt{n}}}-y|$$=$$|({\lambda }_1,\mathrm{\dots },{\lambda }_n){\displaystyle \frac{\epsilon }{2\sqrt{n}}}|$

$=$${\displaystyle \frac{\epsilon }{2\sqrt{n}}}\sqrt{{\lambda }_1^2+\mathrm{\cdots }+{\lambda }_n^2}$

$\le$

so $D\subset B(y,ϵ)\subset U$, and the claim follows. $\mathrm{\square }$