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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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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