$(X, T)$ is an infinite topological space, and $C:= ${{$x$}:{$x$}$∈T$}. $C$ is finite. Does $Y:=X\setminus C$ have any open singleton if we consider the subspace topology on $Y$?
What if X is Hausdorff?
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Do you know what is a subspace topology? – Peter Franek Jun 28 '14 at 05:11
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1{B$\bigcap$Y:B$\in$T} – user136592 Jun 28 '14 at 05:13
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What if B is open in X, B consisting of some x$\in$ C and a y $\notin$ C? – user136592 Jun 28 '14 at 05:17
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It can have. Concider $X={1/n:,n\in\mathbb{N}}\cup{0}$. – Peter Franek Jun 28 '14 at 05:20
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And what is the topology here? – user136592 Jun 28 '14 at 05:25
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Sorry, I thought that $C$ can be infinite. If it is finite, then you just removed an open and closed set $C$ and $Y\subseteq X$ is open. A set is open in $Y$ iff it is open in $X$ (and contained in $Y$). So there are no more singletons remaining. – Peter Franek Jun 28 '14 at 05:28
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I see, but I actually made a mistake, what I do want to put there is C is finite. However, I think I could find an answer by fixing your example a little bit.. Sorry I made the confusion. – user136592 Jun 28 '14 at 05:32
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Yes, exactly, thank you very much! – user136592 Jun 28 '14 at 05:34
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I don't think that question makes sense as written. Clearly $C$ is not a subset of $X$. (It is a subset of $P(X)$.) – Martin Sleziak Jun 28 '14 at 09:24
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1I think the OP probably meant $C={x\in X\mid {x}\in T}$. – Dan Rust Jun 28 '14 at 09:30
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@PeterFranek you do assume then that the space is $T_1$. – Henno Brandsma Jun 28 '14 at 10:24
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@Hemno Brandsma: yes, good point! The counterexample is $X={x,y}$ and $T={\emptyset, {x}, {x,y}}$. – Peter Franek Jun 28 '14 at 10:38
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1@PeterFranek I have an infinite version in my answer, as the OP asked for an infinite space.. – Henno Brandsma Jun 28 '14 at 11:53
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It can, in general.
Let $X$ be $\mathbb{N}$ with the lower topology: all sets $L_k = \{n \in \mathbb{N}: n \le k \}$ for $k \in \mathbb{N}$, plus the emptyset and $X$. Then $0$ is the only isolated point, but in $X \setminus \{0\}$ we have a new isolated point $1$, (it's homeomorphic to $X$ again!).
If $X$ is $T_1$ (so Hausdorff will also do), then finite sets are closed and then $X \setminus C$ is then a (closed-and-)open subset of $X$, so any isolated point of $X \setminus C$ would also have been one of $X$ (as "open in open is open"), which cannot be.
So for $X$ $T_1$ the assertion holds, for mere $T_0$ spaces like the lower topology, this need not hold.
Henno Brandsma
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