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](https://math.stackexchange.com/questions/653106/is-the-cartesian-product-of-two-open-sets-open#:%7E:text=Yes%2C%20the%20product%20of%20two%20open%20sets%20is%20open).

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?](https://math.stackexchange.com/questions/479777/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.