Show that A=$\{(x_1,...x_n) \in \Bbb R | -1\le x_1\le x_2\le ...x_n\le 1\} \subset \Bbb R^n $ is closed. The full question was:
Show that A=$\{(x_1,...x_n) \in \Bbb R \mid -1\le x_1\le x_2\le ...x_n\le 1\} \subset \Bbb R^n       $ is compact, but I was able to show correctly that it is bounded.
However my proof of it being closed didn't hold up.
I tried to show $ A^c  $ open.
So for some $x\in A^c, y\in A$, either


*

*$x_{i+k}<x_i     $ for some $i$, and k is a positive integer.


*

*$\Rightarrow d(x,y)\ge \frac1{\sqrt2} |x_{i+k} - x_i|\ge0\Rightarrow B(x,\frac12 |x_{i+k} - x_i|) \subset A^c  $


*or $\exists\ x_i \notin\ [-1, 1] $


*

*$\Rightarrow d(x,y)\ge |x_i - y_i|\gt 0 \Rightarrow B(x, \left(\frac{|x_i - y_i|}2\right) )\subset A^c$



Any suggestions? The corrector said 'You have only put an open interval around one co-ordinate, you need to do it for all of them' but i don't understand what he meant.
Thanks in advance!
 A: Your corrector was partially incorrect.  The point is that if some $\mathbf{x} = \langle x_1 , \ldots , x_n \rangle$ does not belong to $A$, this is witnessed by a failure of the condition for at most two coordinates of $\mathbf{x}$.  We can then choose open intervals for these coordinates such that regardless of how we pick points to "pass through" these intervals, we get the same failure of being an element of $A$.  This would be the same as taking the projections onto these other coordinates to be the full $\mathbb{R}$.
Let's work through this question a bit more fully. If $\mathbf{x}$ is not in $A$, there are three possibilities.


*

*If $x_1 < -1$, then consider the set $U = ( - \infty , -1 ) \times \mathbb{R} \times \cdots \mathbb{R}$.  This is open in $\mathbb{R}^n$, contains $\mathbf{x}$, and is disjoint from $A$ (since for any $\mathbf{y} = \langle y_1 , \ldots , y_n \rangle \in U$ we must have $y_1 < -1$

*If $x_n > 1$, then consider $U = \mathbb{R} \times \cdots \times \mathbb{R} \times ( 1 , + \infty )$.  again this is open in $\mathbb{R}^n$, contains $\mathbf{x}$, and is disjoint from $A$ (since for any $\mathbf{y} = \langle y_1 , \ldots , y_n \rangle \in U$ we must have $y_n > 1$).

*If $i < n$ is such that $x_i > x_{i+1}$, then let $a = \frac{x_i + x_{i+1}}2$, and consider the set $U = \mathbb{R} \times \cdots \times \mathbb{R} \times ( a , + \infty ) \times ( -\infty , a ) \times \mathbb{R} \times \cdots \mathbb{R}$ (where the "$(a,+\infty)$" appears on the $i^\text{th}$ coordinate).  This is open, contains $\mathbf{x}$, and is disjoint from $A$ (since for any $\mathbf{y} = \langle y_1 , \ldots , y_n \rangle \in U$ we must have $y_i > a > y_{i+1}$).
As you can see, in this third case we needed to consider open intervals for two coordinates.
A: 
Edit: I misread the problem, my apologies. Here's an edited answer
If $\xi_k=(x^1_k,...,x^n_k) \in A$ for all $k$ and $\xi_k$ converges to $\xi=(x^1,...,x^n),$ then $x^i_k$ converges to $x^i$ for every $1 \leq i \leq n.$
For each $k$, $x^i_k \leq x^{i+1}_k$ for each $i$ implies that $\lim_{k \to \infty}x^i_k \leq \lim_{k \to \infty}x^{i+1}_k.$ In particular, this implies that
$$-1 \leq x^1 \leq \cdots \leq x^n \leq 1,$$ which exactly means that $\xi \in A.$
