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 $U+V=\{(x,y)\in\mathbb{R}^2| x>0\}$.

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  • $\begingroup$ 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 $(c_n,d_n)_{n \in \mathbb N}$? $\endgroup$ – LAD Sep 25 '19 at 21:01
  • $\begingroup$ 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. $\endgroup$ – TomTom314 Sep 25 '19 at 21:06

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