Prove that the set $\{(x,y)\in \mathbb{R}^2 : x+y\geq0\}$ is closed using sequences.

Question

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?

Answer 1

Hint: inequalities are preserved by limits.

Answer 2

Let $a = (x,y)$ is an accumulation point of the set. Then there exists a sequence $(a_n) \in \mathbb{R}^{2}$ converging to it. Assume that $a = (x,y)$ isn't in the set. Then we have that $x+y = N < 0$. Now take a ball with radius $\frac{|N|}{2}$ around $a$. Obviously that the intersection with the set is empty, but this is impossible, as $a$ is a limit point of a sequence in the set. Hence, by contradiction $a$ is in the set. Therefore as the set contains all of it's accumulation points it's closed.

Answer 3

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: $\Bbb R^2\setminus f^{-1}(F)=f^{-1}(\Bbb R\setminus F$)]

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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.

Answer 4

We know that $z_n=(x_n,y_n) \to (x,y) \in \Bbb{R}^2 \Leftrightarrow x_n \to^{d_2} x$ and $y_n \to y$ where $d_2$ is the euclideian distance.

Thus let $z_n \in \{(x,y):x+y \geq 0\}$ and $z_n \to (x_0,y_0)$

then $z_n=(x_n,y_n)$ where $x_n \to x_0$ and $y_n \to y_0$ and $x_n+y_n \geq 0$

So $$x_n +y_n \to x_0 +y_0 \Rightarrow x_0+ y_0 \geq 0 \Rightarrow (x_0,y_0) \in \{(x,y):x+y \geq 0\}$$

So the set $\{(x,y): x+y \geq 0\}$ is closed.

Answer 5

Correct me if wrong.

$A:=${$(x,y)\in \mathbb{R^2}: x+y \ge 0$}.

Let $a=(x,y)$ be an accumulation point of the set $A$.

There exists a sequence $(x_n,y_n) \in A$

such that $\lim_{n \rightarrow \infty} (x_n,y_n)= (x,y).$

Since $(x_n,y_n) \in A$:

$x_n+y_n \ge 0.$

Taking the limit:

$ \lim_{n \rightarrow \infty} (x_n+y_n) = x+y \ge 0.$

$\rightarrow$ $(x,y) \in A$, hence $A$ is closed.