Proving that the graph of a function is a closed subset of $\mathbb{R}^{n+m}$

Question

I have a closet subset $A\subset \mathbb{R}^{n} $ and a continuous function $f:A\rightarrow\mathbb{R}^{m} $

Now I would like to prove that $$\text{graph}(f) = \{(x,f(x)):x\in A\}$$

is a closed subset of $\mathbb{R}^{n+m}$. I am not quite sure how to prove this. I suppose if I work with Cauchy-sequences here, this might do the job, but I do not know where to start. Any help would be greatly appreciated!

Answer 1

This is actually quite easy, just note that the function

$$g:\begin{cases}\Bbb R^{n+m}\to \Bbb R^m \\ (x,y)\mapsto y-f(x)\end{cases}$$

Then $\text{graph}(f) = g^{-1}(0)$ is the inverse image of a closed set under the continuous function, $g$, hence closed.

Answer 2

Let $(x_n)$ a sequence of $A$ that converge (let denote $\ell$ its limit). Using continuity, $$\lim_{n\to \infty }(x_n,f(x_n))=(\ell,f(\ell))\in Graph(f),$$ therefore it's closed.