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

Theorem Suppose $\mathbb{R}^{n}$ 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 $\mathbb{R}^{n}$, there exist real numbers $a_{i},b_{i}$ for $i=1,\ldots,n$ such that $a_{i} and $[a_{1},b_{1}]\times\cdots\times[a_{n},b_{n}]$ 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 $\varepsilon>0$ such that $x$ is contained in the open ball $B(y,\varepsilon)$. Let us now set $a_{i}=y_{i}-\frac{\varepsilon}{2\sqrt{n}}$ and $b_{i}=y_{i}+\frac{\varepsilon}{2\sqrt{n}}$ for all $i=1,\ldots,n$. Then $D=[a_{1},b_{1}]\times\cdots\times[a_{n},b_{n}]$ can be parametrized as

 $D=\{y+(\lambda_{1},\ldots,\lambda_{n})\frac{\varepsilon}{2\sqrt{n}}\mid\lambda% _{i}\in[-1,1]\,\mbox{for all}\,i=1,\ldots,n\}.$

For an arbitrary point in $D$, we have

 $\displaystyle|y+(\lambda_{1},\ldots,\lambda_{n})\frac{\varepsilon}{2\sqrt{n}}-y|$ $\displaystyle=$ $\displaystyle|(\lambda_{1},\ldots,\lambda_{n})\frac{\varepsilon}{2\sqrt{n}}|$ $\displaystyle=$ $\displaystyle\frac{\varepsilon}{2\sqrt{n}}\sqrt{\lambda_{1}^{2}+\cdots+\lambda% _{n}^{2}}$ $\displaystyle\leq$ $\displaystyle\frac{\varepsilon}{2}<\varepsilon,$

so $D\subset B(y,\epsilon)\subset U$, and the claim follows. $\Box$

Title open set in $\mathbb{R}^{n}$ contains an open rectangle  OpenSetInmathbbRnContainsAnOpenRectangle 2013-03-22 14:07:46 2013-03-22 14:07:46 matte (1858) matte (1858) 5 matte (1858) Theorem msc 54E35 Interval   