Q:

Problem III.2.7 in textbook Analysis I by Amann: A function induces a topological space

I'm doing Problem III.2.7 from textbook Analysis I by Amann.

Here is my attempt:
Assume $A\subseteq B$, then $B = A \cup (B-A)$. So $h(B)= h(A \cup (B-A)) = h(A) \cup h(B-A)$ by (iii). Hence $h(A) \subseteq h(B)$ and thus $h$ is increasing.
We have $h(\emptyset) = \emptyset$, so $\emptyset^c = X \in \mathcal T_h$. We have $X \subseteq h(X) \subseteq X$, so $h(X)=X$ and thus $X^c = \emptyset \in \mathcal T_h$.
Assume that $(A_i)_{i\in I}$ is a family of sets such that $A_i \in \mathcal T_h$ for all $i \in I$, so $f(A_i^c) = A_i^c$ for all $i \in I$. Let $A = \cup_{i\in I} A_i$. Then $A^c = (\cup_{i\in I} A_i^c)^c = \cap_{i\in I} A_i^c$. We will prove $A \in \mathcal T_h$ by showing $f(A^c) = A^c$. Because $A^c \subseteq f(A^c)$, it suffices to show that $f(A^c)\subseteq A^c$. We have $f(A^c) = f(\cap_{i\in I} A_i^c) \subseteq f(A_i^c) = A_i^c$ because $\cap_{i\in I} A_i^c \subseteq A_i^c$ for all $i \in I$. As such, $f(A^c) \subseteq \cap_{i\in I} A_i^c = A^c$.
Assume that $(A_i)_{i\in I}$ is a finite family of sets such that $A_i \in \mathcal T_h$ for all $i \in I$, so $f(A_i^c) = A_i^c$ for all $i \in I$. Let $A = \cap_{i\in I} A_i$. Then $A^c = (\cap_{i\in I} A_i)^c = \cup_{i\in I} A_i^c$. We have $f(A^c) = f(\cup_{i\in I} A_i^c) = \cup_{i\in I}f( A_i^c) = \cup_{i\in I} A_i^c = A^c$. Hence $A \in \mathcal T_h$.

My questions:

Could you please verify if my proof look fine or contains logical gaps/errors? Any suggestion is greatly appreciated.
From (iii), I get $h(A \cup B)=h(A) \cup h(B)$ for all $A, B \in \mathcal{P}(X)$. As such, $f(\cup_{i\in I} A_i^c) = \cup_{i\in I}f( A_i^c)$ in case $I$ is finite. I feel that it still holds in case $I$ is infinite or uncountable by transfinite induction. I would like to ask of this understanding of me is correct.

Thank you so much!

A:

You cannot use transfinite induction because (iii) gives you enough to prove it for any natural $n$, but not to prove it for $\omega$.
The proof is more than fine, is clear and does not skip any steps. The only issue is that you use $f$ instead of using $h$.
And for (b), as the topology, the collection of open sets, is given by $\mathcal T_h$, the set of the complements of the fixed points of $h$, then the closed sets are the fixed points of $h$. Thus $h(X)=\bar X = \bigcap\{Y\mid$ is closed with $Y\supset X \}$, the closre of $X$.

