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## Cantor type set?

A $proper$ $subinterval$ of an interval $J$ is a subset and interval each of whose endpoints does not coincide with end points of $J$. Let $A_{0}=[0,1]$ be a subspace of $\mathbb{R}$. Let $A_{1}$ be a subspace of $\mathbb{R}$ obtained by removing a proper open interval $I$ in $[0,1]$. Let $A_{2}$ be the one obtained from $A_{1}$ by removing proper open interval $I_{2}^{+}$ and $I_{2}^{-}$ in thecomponents of $A_{1}$. Define $A_{n}$ by removing from $A_{n-1}$ one proper open interval from each of components of $A_{n-1}$. Let $C=\bigcap_{n \in \mathbb{N}}A_{n}$ and give it the subspace topology from $\mathbb{R}$. I would like to know if the set $C$ does have any isolated point. I don't think it's true. But could anyone provide a proof?

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Your first sentence doesn't make any sense. Each of your $A_n$, if I guess rightly, are homeomorphic to the usual sets used to define the Cantor set, and the Cantor set is then a special case of $C$. Have you checked whether it has any isolated point? This should be a standard point-set topology question. – David Roberts Apr 18 2011 at 6:57
The end of the first sentence should be, "endpoints of $J$." I think this question would find a better home at math.stackexchange.com or one of the other sites suggested in the FAQ. – S. Carnahan Apr 18 2011 at 7:03
Sorry for the typo and slow typing. I edited my question. But The set C is not necessarily homeomorphic to the cantor set. For example, I could leave $[0.1/2]$ in each step when I remove the proper open interval. – kchoi Apr 18 2011 at 7:03
I meant that I could leave the closed interval $[0, 1/2]$ in each step when I remove the open intervals. – kchoi Apr 18 2011 at 7:09
Your question reduces down (more or less) to the following: if $x\in C\cap [0,1])$, show that there is a monotone decreasing sequence $(x_n)$ in $C$ with $x_n\to x$. I think you should be able to work things out from here; just keep track of what is going on at each "generation" of the procedure. – Yemon Choi Apr 18 2011 at 7:26
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## 1 Answer

Let me try and provide a proof:

Assume for the sake of contradiction that $x$ is an isolated point of $C$. Notice that the set of endpoints of the intervals removed is contained in $C$. If $x$ is isolated, then it has an $\epsilon$ neighborhood that does not contain any of the points of $C$ except for $x$ and in particular, non of the endpoints of the intervals removed except for maybe itself ($x$ may be such an endpoint itself), but since non of the intervals removed contain $x$ (since $x$ is in $C$) it follows that each such interval is strictly from one side of $x$ and thus at least from one side (if $x$ is an endpoint of some interval itself, then the side that is opposite to that interval), $x$ has a half-open neighborhood contained entirely in $C$ (since it is disjoint from any of the intervals removed) and therefore contradicting the assumption that $x$ is isolated.

I hope it was clear enough (and correct).

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