Compactness exercises from Dugundji I have the following exercises from Topology by Dugundji:

Q1. Let $f\colon X\to Y$ be continuous function and let $\{F_n\colon n\in\mathbb Z^{+}\}$ be a descending of compact subsets of $X.$ Prove that $f[\bigcap_{n=1}^{n=\infty} F_n]=\bigcap_{n=1}^{n=\infty}f[F_n].$


Q2 Let $Y$ be compact and $f\colon Y\to Y$ continuous function. Prove that there exists nonempty set $A\subset A$ such that $f[A]=A$

My attempt for Q1.  Notice that we have $$F_1\supseteq F_2\supseteq\dots$$ of compact subsets of $X$ and we also have $$f[F_1]\supseteq f[F_2]\supseteq\dots\tag{1}$$ Since $f[F_1]$ is compact subset of $Y.$ So, (1) has F.I.P, finite intersection property,  and then  $\bigcap_{n=1}^{\infty}f[F_n]\neq\emptyset.$
Let  $y\in \bigcap_{n=1}^{\infty}f[F_n]$ which implies that  there exists $x\in\bigcap_{n=1}^{\infty} F_n$ such that $f(x)=y.$ Hence, $y\in f[\bigcap_{n=1}^{\infty} F_n].$ We have showed that $$f[\bigcap_{n=1}^{\infty} F_n]\supseteq\bigcap_{n=1}^{\infty}f[F_n].$$ Clearly,  $f[\bigcap_{n=1}^{\infty} F_n]\subseteq\bigcap_{n=1}^{\infty}f[F_n].$This finishes the proof Q1.
My attempt for Q2. We need to construct our set $A$.
Let $A_0=X$ and $A_{n+1}=f[A_n]$ for all $n\geq 0.$ Notcie that $A_n$ is closed and compact for all $n.$ Suppose $A=\bigcap_n A_n$. Notice that $X=A_0\supseteq A1$ and $A_1\supseteq A_2$. So, we have $$A_0\supseteq A_1\supseteq A_2\supseteq\dots $$ Since $X$ is a compact space. So, $A\neq \emptyset .$ Now, $f[A]=f[\bigcap_n A_n]$ and, by Q1, $f[A]=\bigcap_n f[A_n]=\bigcap_n A_n=A$. Since $A$ is closed so this finishes the proof.

Is that right? Can we do question 2 without using question 1?

 A: You say (under attempt Q1)

Let  $y\in \bigcap_{n=1}^{\infty}f[F_n]$ which implies that  there exists $x\in\bigcap_{n=1}^{\infty} F_n$ such that $f(x)=y.$

No, this would follow from $y \in f[\bigcap_{n=1}^{\infty} F_n]$ but this is not given, it's what you have to show (still).
Better attempt: for each $n$ we have that $F'_n:=f^{-1}[\{y\}] \cap F_n \neq \emptyset$, as $y \in f[F_n]$ from the intersection. Now apply the FIP property to those smaller compact sets $F'_n$.
For Q2 just apply Zorn's lemma to the poset of "all non-empty compact $A \subseteq X$ such that $f[A]=A$, ordered by reverse inclusion". For a chain in this poset apply the idea from Q1 to see that the intersection of the chain is an upperbound. A maximal element of this poset is as required. Doing the sequence $A_n$ will work too (but do a better job of showing the decreasingness of the $A_n$ eg by induction), but I like Zorn too. And on general principles, it's nice to use Q1 to do Q2. It's a lemma of sorts, why would you want to avoid it?
