I‘m trying to verify that the sum of two compact operators of Banach spaces is again a compact operator. This question has already been dealt with on this site a few times, but I have a question regarding a step in the proof.
The usual reasoning is as follows:
Let $T, S: V \rightarrow W$ be compact operators, i.e. the closures of $T(B_{\leq 1}(0))$ and $S(B_{\leq 1}(0))$ are compact.
Now since the addition map $\operatorname{add}: W \times W \rightarrow W, (v,w) \mapsto v+w$ is continuous the image of the compact set $\overline{T(B_{\leq 1}(0))} \times \overline{S(B_{\leq 1}(0))}$ is compact and thus is its closure.
My problem with this reasoning is that: $$ \overline{(S+T)(B_{\leq 1}(0))} = \overline{\operatorname{add}(\{(S(v),T(v)) \ : \ v \in B_{\leq 1}(0)\})}\neq \operatorname{add}(\overline{T(B_{\leq 1}(0))} \times \overline{S(B_{\leq 1}(0))}) $$
Or at least I don’t get why, what I wrote as an inequality, should be an equality. So the image of the unit ball under the sum of the operators is not the image of the cartesian product of the addition function. So why is the observation above even helpful?