Let $A\subset\Bbb R^m$ and $B\subset\Bbb R^n$ be closed subsets. Then $A \times B$ is a closed subset of $\Bbb R^{m+n}$ I am preparing for my exam right now and therefore practicing by doing some exercises. I need help for the following task:
Let $A\subset\Bbb R^m$ and $B\subset\Bbb R^n$ be closed subsets. Then $A \times B$ is a closed subset of $\mathbb{R^{m+n}}$
I already found a good answer for a similar task, where $A\subset \mathbb{R}$ and $B\subset\mathbb{R}$ here.
My problem is, that I want to solve this task in a similar way. I want to show that $(A\times B)^c$ is open by doing all steps @DanielFischer did... but the thing here is, that we are working with $\mathbb{R^m}$ and $\mathbb{R^n}$ with m,n also being $\geq2$.
If we have $a\in A\subset \Bbb R$ and $b\in B\subset \Bbb R$, we can get the open balls $B(a,\varepsilon_1)$ and $B(b,\varepsilon_2)$ and then get $B((a,b),\min(\varepsilon_1,\varepsilon_2))\subset B(a,\varepsilon_1)\times B(b,\varepsilon_2)$. And then we want to prove that $(x,y)\in B((a,b),\min(\varepsilon_1,\varepsilon_2))$.
But I don't see how I could write this with $\mathrm a=(a_1,\mathrm,a_n)$ and $\mathrm b=(b_1,\ldots,b_n)$ being vectors. I mean $a(\in \Bbb R)\times b(\in \Bbb R)$ is obviously $(a,b)$. But what is $(a_1,\ldots,a_n)\times (b_1,\ldots,b_n)$
I don't know if its a dumb question but I hope someone of you can help me out.
Also...I found another solution that seems pretty easy but I can't quite follow it:Is product of two closed sets closed?
"$A$ is closed in $X$, so $A^c$ is open, likewise for $B$ in $Y$.
Moreover, $A^c \times Y$ and $X \times B^c$ are both open in $X \times Y$.
Thus $$(A \times B)^c = (A^c \times Y) \cup (X \times B^c)  $$ is open. Hence $A \times B$ is closed."
The problem here is that I can't follow why ($A^c \times Y$) and ($B^c\times X)$ is supposed to be open. Also why is ($A^c \times Y$)$\cup$ ($B^c\times X)$ open? We only learned that the intersection of two opened sets is definetly open. Is there like any proof for my problem?
I am thankful for any advice.
 A: Alternative proof using sequences: Let $((a_k, b_k))_k$ be a sequence in $A \times B$ that converges to $(a,b) \in \mathbb{R}^{m}\times\mathbb{R}^n$. Then $\|a_k -a\| \leq \| (a_k,b_k) - (a,b) \| \to 0$ and thus $\| a_k - a \| \to 0$. Since $(a_k)_k$ is a sequence in $A$ and $A$ is closed, we find $a \in A$. We can apply the same argument and find $b \in B$. In summary: $(a,b) \in A \times B$.
A: We know that $A^c$ is open in $X$ and $B^c$ is open in $Y$. Let's just show that $A^c\times Y$ is open in $X\times Y$: Let $(x,y)\in A^c\times Y$. Since $A^c$ is open in $X$, let $U$ be an open neighborhood of $x$ such that $U\subseteq A^c$. Then, by the definition of the product topology on $X\times Y$, $U\times Y$ is an open neghborhood of $(x,y)$ such that $U\times Y\subseteq A^c\times Y$, so $A^c\times Y$ is open in $X\times Y$.
Also, $(A^c\times Y)\cup(X\times B^c)$ is open because it is the union of two open sets. The fact that an arbitrary union of open sets is open is part of the definition of a topology $\tau$ on some set $S$.
