Let $U$ be open in $\mathbb{R}^2$ and R be an "closed rectangle" in $U$, i.e $R=[a,b]\times[c,d] \subseteq U$ . Then there is an open rectangle $R':=(f,g) \times (h,l)$ such that $R' \subseteq U$ and $R \subseteq R'$.

Can you find a simple proof of this fact?

This seems quite easy to show, but I have only found a quite (unnecessarily?) complicated proof, which goes essentially as follows: First we show that for any line segment $[u,v] \times \{ w\} \subset U$ there is $\epsilon >0$ such that $(u-\epsilon, v + \epsilon)\times (w-\epsilon, w+\epsilon) \subset U$. This can be shown using the compactness of $[u,v] \times \{ w\}$. Once this is shown, we can conclude that $(a-\epsilon, b + \epsilon) \times (c-\epsilon, d + \epsilon)$ by applying the above to line segments of the boundary of $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.