# Cantor type set? [closed]

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 '11 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 '11 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 '11 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 '11 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. –  Captain Oates Apr 18 '11 at 7:26

## closed as off-topic by Ricardo Andrade, David White, Lucia, Andrey Rekalo, Stefan KohlDec 1 at 9:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, David White, Lucia, Andrey Rekalo, Stefan Kohl
If this question can be reworded to fit the rules in the help center, please edit the question.

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.