Im having difficulties trying to prove the following proposition: If $Q$ is reactangle, that is, $Q = [a_1,b_1] \times ... \times [a_n,b_n] $ consisting of all $x$ in $\mathbb{R}^n$ such that $a_i \leq x_i \leq b_i$ for all $i$, then I want to show two things: first that $IntQ = (a_1,b_1) \times ... \times (a_n,b_n)$ and secondly that $Q$ equals the closure of the interior of $Q$. Can someone please help me how can I prove these two statements? Thanks in advance


HINT: For the first part, if $x=\langle x_1,\dots,x_n\rangle\in(a_1,b_1)\times\ldots\times(a_n,b_n)$, let


and show that the open ball of radius $\epsilon$ centred at $x$ is contained within $Q$; this will show that


To finish the job, show that if $x\in Q\setminus\Big((a_1,b_1)\times\ldots\times(a_n,b_n)\Big)$, then there is a $k$ such that $x_k\in\{a_k,b_k\}$. Then use this to show that no open ball centred at $x$ lies entirely within $Q$.

Once you’ve done that, the second part boils down to showing that


Show that each point of $Q\setminus\Big((a_1,b_1)\times\ldots\times(a_n,b_n)\Big)$ is a limit point of $(a_1,b_1)\times\ldots\times(a_n,b_n)$; the observation that these points all have $x_k\in\{a_k,b_k\}$ for some $k$ should help. Then show that no point of $\Bbb R^n\setminus Q$ is a limit point of $(a_1,b_1)\times\ldots\times(a_n,b_n)$; use the fact that if $x\notin Q$, then there is a $k$ such that either $x_k<a_k$ or $x_k>b_k$.

(If you know that products of open sets are open, and products of closed sets are closed, you can shorten the argument considerably. I’m not assuming only that you know that a set is open if and only if it contains an open ball centred at each of its points, and that a set is closed if and only if it’s complement is open.)


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