# Let $C \subseteq X$ be closed, and $D \subseteq X$ be compact. Then $A := \{ x \in X \mid \exists (c \in C, d\in D): x =c-d\}$ is closed.

Good evening, I'm trying to prove this theorem.

Theorem: Let $$X$$ be a normed vector space, $$C$$ be a nonempty closed subset of $$X$$, and $$D$$ be a nonempty compact subset of $$X$$. Then $$A := \{ x \in X \mid \exists (c \in C, d\in D): x =c-d\}$$ is closed.

My questions:

1. Could you please verify if my proof looks fine or contains logical gaps/errors?

2. Does the theorem still hold in case $$D$$ is closed instead of being compact?

Thank you so much for your help!

My attempt:

Let $$(a_n)$$ be a sequence in $$A$$ such that $$a_n \to a\in X$$. By Axiom of Countable Choice, there are two sequences $$(c_n)$$ and $$(d_n)$$ such that $$a_n = c_n - d_n$$ for all $$n \in \mathbb N$$. Since $$D$$ is compact, there is a subsequence $$(d_{n_k})$$ of $$(d_n)$$ such that $$d_{n_k} \to d \in D$$. Then the sequence $$(c_{n_k}) = (a_{n_k} + d_{n_k}) \to (a+d)$$. Because $$C$$ is closed, $$(a+d) \in C$$. It follows from $$a =(a+d) -d$$ that $$a \in A$$. As such, $$A$$ is closed.

"By Axiom of Countable Choice" isn't necessary. It's just by definition of $$A$$. Otherwise, the proof seems fine. You could exchange $$x=c-d$$ by $$x=c+d$$ and write $$A=C-D$$ or $$A=C+D$$.
Counterexample for two closed sets: The set $$\{(x,y)\in\mathbb{R}^2| y\neq 0 \land x\geq y^{-2}\}$$ splits into two components $$U,V$$ with $$U+V=\{(x,y)\in\mathbb{R}^2| x>0\}$$.
• I meant the set $A_n = \{(c,d) \in C\times D \mid c-d = a_n\} \neq \emptyset$ for all $n \in \mathbb N$. Without AC, how can I pick such sequence $(c_n,d_n)_{n \in \mathbb N}$? – LAD Sep 25 '19 at 21:01
• For $a\in X$ you have $a\in A\iff(\exists c\in C, d\in D:a=c-d)$. So if you have a sequence in $(a_n)\subset A$ you know by definiton there is at least one $c_n,d_n$ for each $n$. Therefore $A_n$ can't be emty. – TomTom314 Sep 25 '19 at 21:06