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I can't solve the following problem.

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

please help anyone.

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What have you tried? You'll be more likely to get helpful responses if you show the work you've done, so that people can see what you understand and where you're getting stuck. –  Neal Sep 20 '12 at 11:58
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Could you draw this set if asked? What definition of "closed" are you using? If it's "closed sets have open complements," have you thought about the complement of $A+B$ yet? –  Kevin Carlson Sep 20 '12 at 12:00

1 Answer 1

HINT: Show that $A+B = \{ (x,y) \in \mathbb{R}^2 : |x|\leq 1\}$ then prove that this set is closed by applying the definition of closed point, on every point, or you can look at the complement, that is $\{(x,y)\in\mathbb{R}^2 : |x|>1 \}$ and show that you can find an open ball around every point of that set.

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