Showing that a rectangle is equal to the closure of its interior I'm trying to show that if Q is a rectangle, then Q equals the closure of Int Q. I have that the closure of Int Q is a subset of Q and I'm now working to show that Q is a subset of the closure of Int Q. I understand that this last inclusion is not true in general but I am confused about which property of rectangles I can use to show this. Could someone give me a hint?
 A: 
Definition · A subset $Q$ of $\mathbb{R}^n$ is called a rectangle iff it is of the form
  $$ Q = [a_1,b_1] \times \dotsb \times [a_n,b_n]$$
  for certain $a_1,\dotsc,a_n,b_1,\dotsc,b_n \in \mathbb{R}$ with $a_1 \lt b_1,\dotsc,a_n \lt b_n$.

Let $Q = [a_1,b_1] \times \dotsb \times [a_n,b_n] \subseteq \mathbb{R}^n$ be a rectangle.


*

*We claim $Q^{\circ} = ]a_1,b_1[ \times \dotsb \times ]a_n,b_n[$.
Since $]a_1,b_1[ \times \dotsb \times ]a_n,b_n[$ is an open subset of $Q$, clearly $]a_1,b_1[ \times \dotsb \times ]a_n,b_n[ \subseteq Q^\circ$. We also know $Q^\circ \subseteq Q$.
Now assume $]a_1,b_1[ \times \dotsb \times ]a_n,b_n[ \subsetneq Q^\circ$. Then we had an $x=(x_1,\dotsc,x_n) \in Q^\circ$ with $x_i=a_i$ or $x_i=b_i$ for some $1 \leq i \leq n $. Let $x_i=a_i$ (the other case is very similar). Since $Q^\circ$ is open, there also was an $\epsilon \gt 0$ such that $B_\epsilon(x) \subseteq Q^\circ$. So we had $(x_1,\dotsc,x_i-\epsilon , \dotsc,x_n) \in B_\epsilon(x)$ but — since $x_i \lt a_i $ — also $(x_1,\dotsc,x_i-\epsilon , \dotsc,x_n) \notin Q $ and thus even more $(x_1,\dotsc,x_i-\epsilon , \dotsc,x_n) \notin Q^\circ $, a contradiction. It follows that indeed $Q^{\circ} = ]a_1,b_1[ \times \dotsb \times ]a_n,b_n[$.

*We claim $\overline{Q^\circ} = [a_1,b_1] \times \dotsb \times [a_n,b_n]$.
Since $Q$ is closed and $Q^\circ \subseteq Q$, we clearly have $\overline{Q^\circ} \subseteq \overline{Q} = Q$. Further more we also know $]a_1,b_1[ \times \dotsb \times ]a_n,b_n[ \subseteq \overline{Q^\circ}$.
Now assume $\overline{Q^\circ} \subsetneq Q = [a_1,b_1] \times \dotsb \times [a_n,b_n]$. Then we had an $x=(x_1,\dotsc,x_n) \in Q$ with $x \in \mathbb{R}^n \setminus \overline{Q^\circ}$ and therefore $x_i=a_i$ or $x_i=b_i$ for at least one $1 \leq i \leq n $. Let $I := \{ i \in {1,\dotsc,n} | x_i = a_i \text{ or } x_i = b_i\} \neq \emptyset$. We now continue the proof only for the case $x_j=a_j$ for all $j \in I$ but the other cases can be treated very similar. Since $\overline{Q^\circ}$ is closed, $\mathbb{R}^n \setminus \overline{Q^\circ}$ is open and we have an $ 0 \lt \epsilon$ such that $B_\epsilon(x) \subseteq \mathbb{R}^n \setminus \overline{Q^\circ}$. Let $j \in I$ be arbitrary. Then $ a_j \lt x_j+\epsilon $ and we can choose an $0 \lt \epsilon_j \lt \epsilon$ such that $x_j + \epsilon_j \in ]a_j,b_j[$. For $j \in \{1,\dotsc,n\} \setminus I$ we have $x_j \in ]a_j,b_j[$ anyway. Now consider the point $y=(y_1,\dotsc, y_n)$ with $y_j := x_j + \frac{\epsilon_j}{\sqrt{|I|}}$ for $j \in I$ and $y_j := x_j$ for $j \notin I$. With $k' \in \{ k \in I | \epsilon^2_j \leq \epsilon^2_k \text{ for all } j \in I\}$ (i.e. $\epsilon_{k'}$ has the biggest "$\epsilon$-square") we get
$$ ||y-x|| = \sqrt{\sum_{j \in I} \frac{\epsilon^2_j}{|I|}} \leq \sqrt{|I| · \frac{\epsilon^2_{k'}}{|I|}} = \epsilon_{k'} \lt \epsilon $$
hence $y \in B_\epsilon(x) \subseteq \mathbb{R}^n \setminus \overline{Q^\circ} \subseteq \mathbb{R}^n \setminus Q^\circ$ but also $y \in ]a_1,b_1[ \times \dotsb \times ]a_n,b_n[ = Q^\circ$, a contradiction. It follows that indeed $\overline{Q^\circ} = [a_1,b_1] \times \dotsb \times [a_n,b_n]$.
A: Since proving $[a_1,b_1] \times \dotsb \times [a_n,b_n] \subseteq \overline{Q^\circ}$ was the hardest part in my previous answer, here is another (maybe easier) way to prove this. You need to know that for a set $S \subseteq \mathbb{R}^n $ the closure $\overline{S}$ is exactly the set of all $x \in \mathbb{R}^n $ that appear as limits of arbitrary converging sequences $(s_k)_{k\in\mathbb{N}}$ with $s_k \in S$. Sometimes this property is even the definition of closure in a metric space.
Let $x = (x_1,\dotsc,x_n) \in [a_1,b_1] \times \dotsb \times [a_n,b_n]$ be arbitrary. Define the sets $I := \{i \in \{1,\dotsc,n\} | x_i = a_i \}$ and $ J := \{i\in \{1,\dotsc,n\} | x_i = b_i \}$. For $i \in \{1,\dotsc,n\} \setminus (I \cup J)$ we then have $x_i \in ]a_i,b_i[$. Now define for each $k \in \mathbb{N}$ and each $i \in \{1,\dotsc,n\}$
$$y^{(k)}_i :=\begin{cases}a_i+\frac{b_i-a_i}{2k} &\text{if } i \in I, \\ b_i-\frac{b_i-a_i}{2k} &\text{if } i \in J, \\ x_i &\text{else} \end{cases} $$
and consider the sequence $(y_k)_{k \in \mathbb{N}}$ with $y_k := (y^{(k)}_1,\dotsc,y^{(k)}_n)$. Note that for each $k \in \mathbb{N}$ we have $y_k \in ]a_1,b_1[ \times \dotsb \times ]a_n,b_n[ = Q^\circ$ and
$$\lim_{k \to \infty} y_k = (\lim_{k \to \infty} y^{(k)}_1, \dotsc, \lim_{k \to \infty} y^{(k)}_n) = (x_1,\dotsc,x_n) = x.$$
Therefore, $x \in \overline{Q^\circ}$.
