There is a part of a proof that I am reading that I don't understand.
Let $A$ be a nonempty, compact and convex subset of a topological linear space.
In the proof we construct a commuting family of continuous transformations of $A$ that is closed under composition, say $\mathcal{F^*}$.
We show that the set $\{f(A):f \in \mathcal{F^*}\}$ is a family of closed sets that satisfies the finite intersection property (finite intersections are nonempty).
The proof then says:
Since $A$ is compact we have that $\bigcap_{f \in \mathcal{F^*}}f(A)\neq \emptyset.$
I don't see why this last statement is true. We know that finite intersections are nonempty, but the set $\mathcal{F^*}$ is infinite, so this is an infinite intersection. I don't see how the compactness of $A$ leads to this result. I would appreciate it if someone could explain. Thanks
Fix any $f_0 \in \mathcal F^{*}$. If then intersection is empty then $f_0(A) \subseteq \cup_{f\in \mathcal F^{*}, f \neq f} [f(A)]^{c}$. This gives an open cover for the compact set $f_0(A)$. Hence $f_0(A) \subseteq \cup_{i=1}^{n} [f_i(A)]^{c}$ for some finite collection $(f_i) \subseteq \mathcal F^{*}$. This means $\cap_{i=0}^{n} f_i(A)=\emptyset$, a contradiction.