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Let $X$ be a topological space and let $\left\{ Y_{i}\right\} _{i=1}^{\infty}$ be a descending chain of closed connected subsets of $X$. I know from reading elsewhere that ${\displaystyle \bigcap_{i=1}^{\infty}Y_{i}}$ is not necessarily a connected subspace of $X$ but I have no counter example and I haven't managed to come up with one. There is a counter example here to the same question while also assuming $X$ is compact. However, it uses the quotient topology which I haven't studied about so I would prefer a different counterexample that does not use the quotient topology.

Help would be appreciated!

Eric Wofsey
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Serpahimz
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3 Answers3

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In $\mathbb{R}^{2}$ take two disjoint infinite lines $L_{1},L_{2}$ connected by vertical lines $B_{1},...$ (like a ladder). Then define $$A_{n}=L_{1}\cup L_{2}\cup \bigcup_{k\geq n}B_{k}.$$

These form a descending chain and their intersection will be $L_{1}\cup L_{2}$.

user133100
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Try the following sets: $$Y_i = \bigg([-2,2]\times\mathbb{R}\bigg) - \bigg((-1,1)\times(-i,i)\bigg).$$

That is, $Y_i$ is a closed infinite strip with successively larger open boxes removed. The intersection $$\bigcap_{i=1}^\infty Y_i = ([-2,-1]\cup[1,2])\times\mathbb{R}$$ is disconnected, but each $Y_i$ is closed and connected.

Neal
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    It took me a couple of minute to realise that you we removing a subset of the open disk, not taking a product of the punctured open disk (there really should be some parentheses here). Given that, are the sets you describe not open in $\mathbb{R}^2$? – Dan Rust May 06 '13 at 22:16
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    The example then no longer satisfies the OP's condition. – Dan Rust May 06 '13 at 22:23
  • @DanielRust: Same idea: Remove a "fattened interval" $(-1/2,1/2) \times (-1+1/i, 1-1/i)$ from $[-1,1]^2$. – Martin May 06 '13 at 22:25
  • On the same link to the referred question they prove that no counterexample can be found in normal spaces. – Berci May 06 '13 at 22:27
  • @Martin this still doesn't seem to satisfy the OP's requirement (unless i'm missing something). – Dan Rust May 06 '13 at 22:28
  • @DanielRust: It's a descending chain of closed connected sets, so what's missing? – Martin May 06 '13 at 22:29
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    Like Berci said in the link I gave they prove that the claim is true in a compact hausdorff space (which is also a normal space) and thus true in every metric space. So no subset of $\mathbb{R}^{2}$ (with metric topology) is going to work as a counter example. – Serpahimz May 06 '13 at 22:30
  • @Martin The OP wants their intersection to be not connected. – Dan Rust May 06 '13 at 22:30
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    @Serpahimz: I don't follow your deduction from compact to normal to metric. Replace the closed square $[-1,1]^2$ by the open square $(-1,1)^2$ and remove $(-1/2,1/2) \times (-1+1/i, 1-1/i)$ in my previous example. Then the intersection is $((-1,-1/2] \cup [1/2,1)) \times (-1,1)$ which is manifestly disconnected. – Martin May 06 '13 at 22:36
  • @Martin: Sorry I got things a bit mixed up indeed. The claim is true in a compact Hausdorff space (which in particular is normal) but the proof given in the link doesn't show the claim to be true for a normal space and thus not necessarily true for every metric space. Obviously not every metric space is compact Hausdorff (it is Hausdorff of course). Regarding your example, I'm not sure I get why are the sets you suggested closed in $\left(-1,1\right)^{2}$? – Serpahimz May 06 '13 at 22:55
  • @Serpahimz $\mathbb{R}^2$ is not a compact metric space. – Neal May 06 '13 at 22:56
  • @DanielRust Oh my goodness, I missed the condition that the $Y_i$ be closed. I'm editing a new example into my answer now. – Neal May 06 '13 at 22:57
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    @Neal: Indeed, already corrected myself. $\left[-1,1\right]^{2}$ however is compact Hausdorff so Martin's original counter example was incorrect. Regarding $\left(-1,1\right)^{2}$ like I mentioned in my previous comment I don't immediately see why the sets Martin suggested are closed in $\left(-1,1\right)^{2}$. – Serpahimz May 06 '13 at 22:59
  • @Serpahimz They are the intersection of the closed sets $\mathbb{R}^2 - (\mbox{fattened interval}_i)$ with the space $(-1,1)^2$, so they are closed in the subspace topology. – Neal May 06 '13 at 23:03
  • @Serpahimz: yes, my first example was incorrect (I missed that the intersection still contains ${-1} \times [-1,1]$ and ${1} \times [-1,1]$.) // The second sets are closed since I remove open sets from $(-1,1)^2$ – Martin May 06 '13 at 23:03
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    Neal's new answer and Martin's revised example look to both work fine to me. – Dan Rust May 06 '13 at 23:04
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The answer here can also be said without the quotient:

Let $X:=[-1,1]\cup\{0'\}$ where $0'$ is a new, fictive element, playing the role of a second origin, and define the topology on $[-1,1]$ as usual, and let $(-a,b)\setminus\{0\}\cup\{0'\}$ be a base for open neighborhoods of $0'$.

Now consider $Y_n:=[-1/n,1/n]\cup\{0'\}$.

Community
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Berci
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