Prove that the set $\{(x,y)\in \mathbb{R}^2 : x+y\geq0\}$ is closed using sequences. I can intuitively realize why this is true. All the points that are on the line $y=-x$ are in the set. How can I prove this with sequences?

Hints:


*

*The proposed set is the preimage of $[0,\infty)$ under the continuous map $f(x,y)=x+y$

*$[0,\infty)$ is a closed set.

*If $F\subset\Bbb R$ is closed and $f$ is a continuous map $\Bbb R^2\to\Bbb R$, then $f^{-1}(F)$ is closed (try to prove this one!) [Hint: $\Bbb R^2\setminus f^{-1}(F)=f^{-1}(\Bbb R\setminus F$)]

Alternatively, here's a sequential approach:
Let $A=\{(x_i,y_i)\}_{i\geq 0}\subset F$ which converges to $(x,y)\in\Bbb R^2$. We need to show that $(x,y)\in F$
Now, we have $x_i+y_i\geq 0~\forall~i\geq 0$.
Now, suppose that $x+y=k\lt 0$.
Consider the open ball $B_\epsilon(x,y)$ where $\epsilon\lt -k/2$. Then, note that this ball doesn't contain any element of $A$ since for any $(X,Y)$ in the ball, we have $$X+Y\lt x+y+2\epsilon\leq x+y-k=k-k=0$$
thus $(X,Y)\notin A$ for all $(X,Y)$ of the open ball, contradicting that $A$ converges to $(x,y)$
Thus, our assumption that $x+y\lt 0$ is wrong, hence $x+y\geq 0$, hence $(x,y)\in A$, hence $A$ is closed.
