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This is an old exam question that I don't have a solution to:

Let $X$, a compact Hausdorff (T2) space, and let $\phi$ a family of closed, non-empty, and connected subsets of $X$, such that for every $A, B \in \phi$, $A \subset B$ or $B \subset A$.

Prove that $Y:= \cap \{A: A \in \phi\}$ is connected.

I tried to solve this question with a friend, and this is what we came up with:

  • Obviously, $X$ is normal space (T4).
  • We tried to see what happens if $Y = U_1 \cup U_2$, disjoint and open sets (and?)
  • We tried to use nets and maybe see if we can create a net that converges to $x \neq y$ (and contradict X being Hausdorff space).
  • We tried to work with continuous functions, but this idea didn't lead us anywhere either.

I feel like there is a simple observation that we are missing. Any ideas?

Thanks!

Stahl
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Hila
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  • Hint: The complement of the intersection is the union of the complements. Maybe you can say something about finite subcovers and relate that back to the intersection. –  Feb 21 '13 at 19:24
  • @ShawnHenry Is that really going to do anymore then show that the intersection is non-empty? – JSchlather Feb 21 '13 at 19:30
  • Yup, if you assume that the intersection is not connected and throw that into the mix. –  Feb 21 '13 at 19:35
  • @ShawnHenry I'm still not seeing it, could you provide a sketch? Brian has already provided a mostly complete answer. – JSchlather Feb 21 '13 at 20:30
  • @JSchlater: If Y is empty, it's connected and you're done. Otherwise suppose $Y$ is nonempty and not connected. Then there are open sets U and V such that U∩V=∅ and Y=(U∩Y)∪(V∩Y). Then U, V, and {X−A|A∈ϕ} form an open cover of X. Thus there are finitely many open sets X−Ai for i=1,...n plus U and V that cover X. The intersection of the Ai is just the smallest one, call it A. Then X−A, U, and V cover X, so U∪V contains A. But that contradicts that A is connected. –  Feb 21 '13 at 20:37
  • @ShawnHenry I don't think you can assume there exists a separation such that $U \cap V=\emptyset$ in the total space. It may be true that such a separation exists in the case of a normal space, but it need not be true in general. For instance we could consider $\mathbb R \cup {\ast}$ where the topology is generated by $(a,b)\cup{\ast}$. – JSchlather Feb 21 '13 at 23:11
  • @JSchlather: A compact Hausdorff space is normal. –  Feb 21 '13 at 23:27
  • Of course, Brian's solution essentially includes a proof of this fact, but you can prove it directly without the Tietze theorem. –  Feb 21 '13 at 23:38

2 Answers2

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Re-corrected: $\newcommand{\cl}{\operatorname{cl}}$Suppose that $Y=H\cup K$, where $H\cap K=\varnothing$, $H\ne\varnothing\ne K$, and $H$ and $K$ are clopen in $Y$. Let $f:Y\to\{0,1\}$ take $H$ to $0$ and $K$ to $1$, use the Tietze extension theorem to extend $f$ continuously to $\hat f:X\to[0,1]$, and let

$$\begin{align*} U&=\hat f^{-1}\left[\left[0,\frac12\right)\right]\supseteq H\;,\\ V&=\hat f^{-1}\left[\left(\frac12,1\right]\right]\supseteq K\;,\text{ and}\\ F&=\hat f^{-1}\left[\left\{\frac12\right\}\right]\;. \end{align*}$$

For each $A\in\phi$ we must have $F\cap A\ne\varnothing$, as otherwise $U\cap A$ and $V\cap A$ would be a disconnection of $A$. Thus, $\mathscr{C}=\{F\cap A:A\in\phi\}$ is a nested family of non-empty compact sets. But then $\varnothing\ne\bigcap\mathscr{C}=F\cap Y=\varnothing$, which is absurd.

Brian M. Scott
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  • How do we know that for a point $p$ in $K$, there is an open set $W\ni p$, such that $W\cap U=\emptyset$? – Stefan Hamcke Feb 21 '13 at 19:59
  • @Stefan: Because $V$ has that property. – Brian M. Scott Feb 21 '13 at 20:00
  • I meant: We had to show the above to show that $p$ is in $V$. Cause I'd like to prove the direction $K\subseteq V$. And we just know that there is a neighborhood rel $Y$ about $p$ not intersecting $U$. – Stefan Hamcke Feb 21 '13 at 20:02
  • @Stefan: All you have to prove is that $K\cap\cl U=\varnothing$, which is clear, since $K$ is relatively open in $Y$. – Brian M. Scott Feb 21 '13 at 20:05
  • $K$ is relatively open in $Y$ means that there is an open set $W$ with $W\cap Y=K$ and $W\cap U\cap Y=\emptyset$. But $W$ could intersect $U$ outside of $Y$ – Stefan Hamcke Feb 21 '13 at 20:12
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    @Stefan: Yes, it could, but it doesn’t matter. If you can’t see it any other way, let $f:Y\to{0,1}$ take $H$ to $0$ and $K$ to $1$, use the Tietze extension theorem to extend $f$ to $\hat f:X\to[0,1]$, and let $U=\hat f^{-1}\left[\left[0\frac12\right)\right]$. Actually, I should do that anyway: I’ll change the answer. – Brian M. Scott Feb 21 '13 at 20:21
  • That's great. I completely forgot about normality of $X$. And if I hadn't then I certainly wouldn't have thought about applying the Tietze theorem in this manner. I suppose you mean $V=\hat f^{-1}[(\frac12,1]]$. Thanks alot :-) – Stefan Hamcke Feb 21 '13 at 20:27
  • @Stefan: Thank you, for forcing me to be more careful about defining $U$. – Brian M. Scott Feb 21 '13 at 20:28
  • @BrianM.Scott can you please explain to me why $U\cap A$ and $V\cap A$ would be a disconnection of $A$? I don't see why $A$ is equal to that union. If $x\in A$, why $x\in U$ or $x\in V$? – JonSK Jan 20 '16 at 06:34
  • @JonSK: In fact there was an oversight in what I wrote back then. I’ve just corrected it; see if it makes sense now. – Brian M. Scott Jan 20 '16 at 07:03
  • @BrianM.Scott it does, thank you very much! – JonSK Jan 20 '16 at 16:09
  • @JonSK: You’re very welcome; thanks for catching the problem! – Brian M. Scott Jan 20 '16 at 22:39
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Seems to me that if either $A \subset B$ or $B \subset A$ then for any $ Y = \bigcap [ A \mid A \in \phi ]$ there exists an set in this union, call it $H$, such that for any other set in the union, lets call them $K_\lambda$, $ H \subset K_\lambda$. From which we can conclude that $ Y = H$ and is thusly connected.

Intuitively, there is a minimal element in this union that is induced by the ordering on $\phi$ so this element is our 'limiting factor' so to speak.

Kevin Doherty
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