# Continuous image of the intersection of decreasing sets in a compact space

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.

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You definitely need more strength. Take $X =[0,1], f: X \rightarrow X, x\mapsto \frac{1}{2}, B_{\epsilon } = (0, \epsilon ).$ –  user17794 Jun 27 '12 at 21:36
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. –  Paul Cuff Jun 27 '12 at 22:02
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) –  Paul Cuff Jun 27 '12 at 22:05
@Asaf The original post did not mention closed in the statement of the problem, only as a possible remedy. –  user17794 Jun 27 '12 at 22:12
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. –  Asaf Karagila Jun 27 '12 at 22:30

Suppose that $f\colon X\to Y$, then we assume that $X$ is compact and $f$ is continuous, however we need to assume that $Y$ is $T_1$ or that $X$ is Hausdorff.

First observe that since $B_\epsilon$ is closed it is compact too. Second we observe that the continuous image of a compact set is compact. (These two facts are true for all compact spaces, not only to Hausdorff spaces)

Since we also took the $B_\epsilon$ descending if the intersection is empty then there is $\epsilon$ such that $B_\epsilon=\varnothing$, and so $f(B_\epsilon)=\varnothing$ and the conclusion follows.

If the intersection is non-empty then we are in a situation where $f(B_\epsilon)$ form a decreasing chain of compact non-empty sets. The intersection cannot be empty, since by compactness we would have to have an empty set within the intersected family, which means $f(B_\epsilon)=\varnothing$, in contradiction to our assumption that we are in the case that $B_\epsilon\neq\varnothing$.

Furthermore, it is trivial that $f(B_0)\subseteq\bigcap f(B_\epsilon)$. So we only have to show the other direction. Suppose $y\in\bigcap f(B_\epsilon)$, then $y\in f(B_\epsilon)$ for all $\epsilon$, let $A=f^{-1}(y)$, this is a closed set, let $A_\epsilon=A\cap B_\epsilon$, this is a decreasing family of closed sets in a compact space, so by a similar argument as above we have that the intersection cannot be empty and must be a subset of $B_0$, therefore $y\in f(B_0)$ as wanted.

To see that we have to have $T_1$ in our assumptions, consider $\{0,1\}$ with the topology $\{\varnothing,\{1\},\{0,1\}\}$. Note that $1$ is dense and that $0$ is an accumulation point of $\{1\}$.

Consider the space $X=\mathbb N\cup\{\infty\}$ and the topology generated by initial segments of $\mathbb N$. So $U$ is open in $X$ if and only if $U=\varnothing$ or $U\cap\mathbb N=\{k\in\mathbb N\mid k<n\}$ for some $n\in\mathbb N$, and if $U\neq X$ then $\infty\notin U$.

Observe that $X$ is compact since if we cover $X$ by open sets we had to include $X$ itself in the covering, since the only open set which contains $\infty$ is $X$.

Now let $f\colon X\to\{0,1\}$ be defined as $f(n)=1$ for $n\in\mathbb N$ and $f(\infty)=0$. This is continuous since the preimage of $\{1\}$ is open, and the preimage of $\{0,1\}$ is open.

Let for $B_{\frac1n}=\{\infty\}\cup\{k\geq n\mid k\in\mathbb N\}$. Those are closed sets since their complement is an initial segment of $\mathbb N$. whose intersection is not empty either. Observe that $f(B_n)=\{0,1\}$ for all $n$, however $\bigcap B_{\frac1n}=\{\infty\}$ and $f(\{\infty\})=\{0\}\neq\bigcap f(B_{\frac1n})$.

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For the newbs (me), can you explain why this fails if the space is $T_1$ or the range is $T_0$? I've only learned topology as generalizing from metric spaces, so at lower separations I don't understand as well. –  Eric Stucky Jun 27 '12 at 22:36
@Eric: I'm not sure it fails at all. I wanted to give an initial answer, since these assumptions may be what the OP is dealing with anyway. It's late and I don't know if I can come up with the proof for the general case tonight, though. –  Asaf Karagila Jun 27 '12 at 22:39
@Eric: I tried to give a counterexample. It wasn't easy, but it was fun! :-) –  Asaf Karagila Jun 27 '12 at 23:27