# Arzela-Ascoli and adjoint of compact operator compact

I have seen in this thread a nice answer where it is shown that Thread

that the adjoint operator of a compact operator is compact by using the Arzela Ascoli theorem. Unfortunately, there is one thing I do not understand: It is not shown that $(f_n)_n$ is closed, so why is this sequence compact then? It is only shown that it is equicontinuous and bounded but not closed and since we derive from Arzela Ascoli that there is a convergent subsequence we probably need compactness.

I should point how we formulated Arzela Ascoli: Let $M \subset C(S)$ where $S$ is a compact set, then $M$ is compact iff: $M$ is bounded, closed and equicontinuous.

Thank you in advance for every helpful comment.

## 1 Answer

If you let $\mathcal{F} = \{ f_n | n\in \mathbb{N} \}$ then this set is not necessarily closed, as you say. However, it has compact-closure and so the sequence $f_n$ converges to a point in $\overline{\mathcal{F}}$ though not necessarily to a point in $\mathcal{F}$ itself.