4
$\begingroup$

Suppose $B_{\epsilon}$ are closed subsets of a compact space and $B_{\epsilon} \supset B_{\epsilon'} \quad \forall \epsilon > \epsilon'$. Furthermore, $B_0 = \bigcap_{\epsilon>0} B_{\epsilon}$. For a continuous function $f$ can we conclude that $$f(B_0) = \bigcap_{\epsilon>0} f(B_{\epsilon})?$$

I believe the answer to be yes. It seems this should be a well-known property---I'm having trouble finding a reference.

  • 0
    You definitely need more strength. Take $X =[0,1], f: X \rightarrow X, x\mapsto \frac{1}{2}, B_{\epsilon } = (0, \epsilon ).$2012-06-27
  • 0
    Hmm. I'm wondering whether to state that $B_0 \neq \emptyset$ or that $B_{\epsilon}$ are closed. Either is true in the case I am considering. Probably the closed condition will get the job done.2012-06-27
  • 0
    On second thought, I see that $B_0 \neq \emptyset$ is not good enough. We could easily extend your example so that some island vanishes in the intersection even though the intersection is non-empty. (Makes me wonder whether connected could remedy that, but that's beside the point)2012-06-27
  • 0
    @Asaf The original post did not mention closed in the statement of the problem, only as a possible remedy.2012-06-27
  • 0
    Paul, I posted an answer under the assumption that $f\colon X\to Y$ and $X$ is a compact Hausdorff space; and $Y$ is $T_1$. If your assumptions on $X$ and $Y$ are different please add this to the question.2012-06-27
  • 0
    Thank you very much. The problem was more nuanced than I thought. At least in Euclidean space everything works fine.2012-06-28

1 Answers 1