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

Theorem Suppose ${ℝ}^n$ is equipped with the usual topology induced by the open balls of the Euclidean metric. Then, if $U$ is a non-empty open set in ${ℝ}^n$, there exist 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 in some open set. Since the topology has a basis consisting of open 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 be parametrized 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. $\mathrm{\square }$

Title	open set in ${ℝ}^n$ contains an open rectangle
Canonical name	OpenSetInmathbbRnContainsAnOpenRectangle
Date of creation	2013-03-22 14:07:46
Last modified on	2013-03-22 14:07:46
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	5
Author	matte (1858)
Entry type	Theorem
Classification	msc 54E35
Related topic	Interval