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Let $A=\{(x,y)\in \mathbb R^2:\max\{|y|,|x|\}\leq 1\}$ and $B=\{(0,y)\in \mathbb R^2:y\in \mathbb R\}$. Show that $A+B$ is a closed subset of $\mathbb R^2$

My try:

let $z_n=x_n+y_n$ be a sequence in $A+B$ converging to $z\in \mathbb R^2$ .To show $z\in A+B$. Now $x_n\in A$

$\implies x_n=(a_n,b_n) $

where $|a_n|,|b_n|\leq 1$and $y_n=(0,c_n)$. Can you help me to proceed further please

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$A$ is compact and $B$ is closed. Now it is a general fact that (in Euclidean spaces) in this case $A+B$ is closed (try to prove this before reading further).

In fact, take any sequence $p_n=a_n+b_n\to p\in\mathbb{R}^2$, where $a_n\in A$ and $b_n\in B$. We have to prove that $p\in A+B$.
As $A$ is compact, upon possibly passing to a subsequence, we have $a_n\to a\in A$.
So $b_n=p_n-a_n\to p-a$ and since $B$ is closed we have $p-a\in B$.
Call $b:=p-a$; we just proved that $p=a+b$ with $a\in A$ and $b\in B$, so we are done.

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Despite the fact that there is a more general result mentioned in another answer, in this specific case it might be easier to show that $$A+B=\{(x,y)\in\mathbb R^2; |x|\le 1\}$$ and then show that this set is closed.

If you try to draw a picture, you should see almost immediately what $A+B$ is, but let us try to do also a formal proof.

Let $C=\{(x,y)\in\mathbb R^2; |x|\le 1\}$. It is easy to see $A+B\subseteq C$.

On the other hand if $(x,y)$ is arbitrary point from $C$, i.e., if $|x|\le 1$, you wan to express it as the sum $$(x,y)=(x,y_1)+(0,y_2)$$ where $|y_1|\le1$ and $y_2$ is arbitrary. Finding such $y_{1,2}$ should be rather straightforward, too.

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