Here is part of a proof that I don't understand:
Let $X$ be a topological linear space and $K \subset X$ be non empty, compact and convex.
We have $\mathcal{F^*}$ is a commuting family of continuous transformations on $K$ that is closed under composition.
Then for all $f \in \mathcal{F^*}$ the set $f(K)$ is closed.
I don't see why this is true. At first I assumed it was due to the continuity of $f$ preserving the closedness of $K$, but we don't know that $K$ is closed, only that it is compact and convex, which as far as I know does not imlply that $K$ is closed in general.