0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ I'm just really tired. I've deleted my answer and that's that. $\endgroup$ – Asaf Karagila Oct 24 '14 at 18:49
0
$\begingroup$

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 $\bigcap_{n \geq 0} f[C_n] = \{0, 1\} \neq f[C] = \{0\}$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – elchaltendude Oct 25 '14 at 3:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.