4

I'm trying to prove below theorem. My proof is much simpler than this one. I'm afraid that I made some subtle mistake. Could you have a check on it?

Let $E$ be a topological vector space, and $A, B \subseteq E$ with $A$ compact and $B$ closed. Then $C :=A+B$ is closed.

My attempt: Let $(c_d)_{d\in D}$ be a net in $C$ that converges to $c\in E$. By axiom of choice, we can write $c_d = a_d+b_d$ for some $a_d \in A$ and $b_d \in B$. Because $A$ is compact, the net $(a_d)_{d\in D}$ has a convergent subnet $(a_{\psi (d)})_{d \in D}$ such that $a_{\psi (d)} \to a$ for some $a \in A$. By definition of net convergence, $c_{\psi (d)} \to c$. Then $b_{\psi (d)} = c_{\psi (d)} - a_{\psi (d)} \to c-a$. Because $B$ is closed, $c-a \in B$ and thus $c = a + (c-a) \in A + B= C$. This completes the proof.

Akira
  • 17,367

1 Answers1

2

To me, it is totally fine and it is a nice generalization to spaces that are not sequential spaces (for which the topology is not fully described by sequences). The proof is essentially the same (with some remarks like the use of the axiom of choice). As another reference I post here Duchamp Gérard H. E.’s answer that is very similar to your (correct) work: https://math.stackexchange.com/a/2079363/865323

Son Gohan
  • 4,397