I am looking for a 'direct' way to show the following statement:
Problem: Let $V$ be a normed vectorspace, show that if $A$ is compact and $B$ is closed then $A+B:= \lbrace a+b \mid a \in A, b \in B \rbrace$ is closed
My approach: I am trying to show this statement directly i.e. premise $\implies$ conclusion. This was ridiculously easy for $A,B$ compact $\implies A+B$ compact. I don't seem to have much luck with this one though.
Let $(x_n)$ be a convergent sequence in $A+B$ such that $x_n \to x$. If I manage to show that $x \in A+B$ then I am done.
It is true that $x_n = a_n + b_n$ for all $n \in \mathbb{N}$ and $(a_n,b_n) \in A \times B$. Luckily I have that $A$ is compact, that means that there exists a subsequence $(a_{n_k})$ of $(a_n)$ such that $a_{n_k} \to a \in A$ thanks to compact $\equiv $ bound & closed and Bolzano-Weierstrass.
Now I fail to make any statement about the sequence $(b_n) \in B$. I know that if $(b_n)$ converges, then thanks to $B$ being closed it follows that the limiting point would also be a member of $B$. However it feels rather vague to me to say that $(b_n)$ converges because $(x_n) \in A + B$ converges.
Could someone give me some nudges in the right direction? Or do I need to forfeit the direct approach and try to come up with a contradiction if I assume $x \notin A+B$? This seems to be much harder.