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Statement of the problem:

Let $E$ be a Normed Vector Space over the real numbers. Let $A, B$ be subsets of $E$ such that:

$A$ and $B$ are non-empty, $A \cap B = \emptyset $. Assume $A$ is closed and $B$ is compact. Define the set $C := A - B = \cup_{b \in B} (A - b)$. Here, $A - b = \{ a - b : a \in A \}$. i.e. a translation of the set $A$ in the direction of $-b$.

Prove that C is closed.

I am sure that $B$ being compact is key here, and suspect I should consider finite covers in some way, but I cannot figure out why $A - S$ should be closed for any potentially infinite set $S$... Could someone please offer me a hint? Thank you.

gbnhgbnhg
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1 Answers1

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Assume $z$ is a point from the boundary of $A-B$. There exists a sequence $z_n\in A-B$ for which $z_n\to z$. Here, $z_n=a_n-b_n$, and $a_n\in A, b_n\in B$. But $A$ is a compact set, so there exists a convergent subsequence $a_{n_k}\to a\in A$. As a hint, try to prove that $z$ also belongs to $A-B$.

MGy
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