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My intuition tells me that this doesnot hold becuase $A+F = \cup_{a \in A}(a+F)$ and arbitrary union of closed sets is not closed.

I was thinking of the following example $A = \{(\frac{1}{n},0):n\ge 1\}\cup \{(0,0)\} \cup \{(0,\frac{1}{n}):n\ge 1\}$ .The set $A$ is closed in $\mathbb{R}^2 - A$ under euclidean metric.

Let $F = [a,b] \times [c,d]$ (This set $F$ is closed in $\mathbb{R}^2$ under euclidean metric).

However I have a feeling that $A+F$ is not closed.Is this example correct?If not can someone modify this example..

Edit 1:This are the follow up questions that I was trying to think of :

1.Is there a way to write $(a,b] \times (c,d] $ or $(a,b) \times \{0\}$ as the sum of closed spaces $A,F$?

2.$ A + F$ where $A = \{(a-n,0):n \ge 1\}$ and $F = \{(a+n-\frac{1}{n},0):n\ge 1\}$ . $(a,0)$ does not belong to $A+F$ so it is not closed but $A,F$ is closed.

3.What about $A-F$?

ThirstForMaths
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1 Answers1

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Let $A= \{(x, 1-e^{-x}) | x \ge 0 \}$ and $F= \{(x, 1-e^{x}) | x \le 0 \}$.

Then $\{0\} \times [0,2) \subset A+F$ but $(0,2) \notin A+F$.

copper.hat
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