# Intersection of the image of a decreasing chain of sets

Just a quick question that goes as follows.

Let $X$ be a nonempty set, $\mathcal{C}$ a decreasing chain of subsets of $X$ with nonempty intersection and $f: X \rightarrow X$ a function such that $f(C) \subset C$ for every $C \in \mathcal{C}$. Is it true that $$\bigcap f(\mathcal{C}) \subset f\left(\bigcap \mathcal{C}\right)$$

Seems to be pretty obvious for finite and countable chains, but just not sure about the general case...

• I'm just really tired. I've deleted my answer and that's that. – Asaf Karagila Oct 24 '14 at 18:49

Counterexample: Let $f: \omega \rightarrow \omega$ be defined by $f(0) = f(1) = 0$ and $f(n) = 1$ if $n \geq 2$. For $n \geq 0$, let $C_n = \{0, 1\} \bigcup \{n+2, n+3, \dots\}$. Then $C_n$'s are decreasing and $f[C_n] = \{0, 1\} \subseteq C_n$ for each $n$. Let $C = \bigcap_{n \geq 0} C_n = \{0, 1\}$. Then $\bigcap_{n \geq 0} f[C_n] = \{0, 1\} \neq f[C] = \{0\}$.
• Thanks for your help. I should probably explain where this comes from: suppose we have a nonempty set $X$ and a function $f: X \rightarrow X$. Let $\mathcal{F}$ be a family of subsets of $X$ such that (a) $X \in \mathcal{F}$, (b) if $A \in \mathcal{F}$ then $f(A) \in \mathcal{F}$, (c) if $\mathcal{C}$ is a chain in $\mathcal{F}$, then $\bigcap \mathcal{C} \in \mathcal{F}$, and (d) if $A \in \mathcal{F}$ then $f(A) \subset A$. Define then a new function $F: \mathcal{F} \rightarrow \mathcal{F}$ by $F(A) = f(A)$. I've been wondering if $F$ is chain-continuous. – elchaltendude Oct 25 '14 at 3:35