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If we have a topological space $X$, consider the following:

Let $p \in X$. Then $\{p\}$ doesn't have any limit points because it's a single subset of $X$, so $\{p\}$ is closed.

Is this correct? So every single point subset in a given topological space is closed? Where are my mistakes?

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In a metric topology every singleton set is closed. There are examples of non-metric topologies where this is not true, but they are ruled by usual regularity assumptions such as assuming a Haussdorf space. –  kjetil b halvorsen Sep 28 '12 at 2:14
thank you for the reply, I know that but I can't see where I'm wrong in my reasoning. –  user42912 Sep 28 '12 at 2:16
Does a singleton contain any other point in its neighborhood? I don't think so... –  Don Larynx Oct 12 '13 at 21:53
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2 Answers

up vote 4 down vote accepted

You might find the following useful:

A topological space $X$ is a $T_1$-space if whenever $x,y\in X$ with $x\ne y$, there is an open set $U$ such that $x\notin U$ and $y\in U$.

Proposition. $X$ is $T_1$ iff $\{x\}$ is closed for each $x\in X$.

Proof. Suppose first that $X$ is $T_1$, and let $x\in X$. They for each $y\in X$ there is an open $U_y\subseteq X$ such that $y\in U_y$ and $x\notin U_y$. Let $U=\bigcup_{y\in X\setminus\{x\}}U_y$; then $U$ is open, and $U=X\setminus\{x\}$, so $\{x\}$ is closed. Now suppose that $x\in X$ and $\{x\}$ is closed. Then $U=X\setminus\{x\}$ is open, and if $x\ne y\in X$, then $U$ is an open set such that $x\notin U$ and $y\in U$. $\dashv$

There are many spaces that are not $T_1$. For example, let $X=\Bbb N$, for each $n\in\Bbb N$ let $$U_n=\{k\in\Bbb N:k<n\}\;,$$ and let $\tau=\{\Bbb N\}\cup\{U_n:n\in\Bbb N\}$; then $\langle X,\tau\rangle$ is a space that isn’t $T_1$. In fact, if $m,n\in X$ with $m,n$, and $U$ is any open set containing $n$, then $m\in U$ as well. Even simpler is the Sierpiński space, whose underlying set is $\{0,1\}$ and whose open sets are $\varnothing,\{1\}$, and $\{0,1\}$: there is no open set containing $0$ but not $1$.

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So basically we can interpret $U_y \subseteq X$ as a set that doesn't exist, correct? –  Don Larynx Oct 12 '13 at 21:59
@Don: As it stands, that really doesn’t make sense. There are spaces $X$ containing points $x$ and $y$ such that every open nbhd of $y$ contains $x$, if that’s what you mean, but the symbol $U_y$ has no standard meaning, so it makes no sense to say that in some space it does or doesn’t exist. –  Brian M. Scott Oct 13 '13 at 6:32
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Let $X=\{1,2,3\}$. Let $\mathcal{T} = \{ \emptyset, \{1,2\}, \{1,2,3\} \} $. Then the set $\{ 2 \}$ is not closed. It is not open either.

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yes, I know we have counter examples, that's why I would like to know which part of my proof I made mistakes. –  user42912 Sep 28 '12 at 2:30
"...$\{p\}$ doesnt have any limit points.." this is where you have a problem. In Scott's example $3$ is a limit point of $\{2\}$. –  Mustafa Gokhan Benli Sep 28 '12 at 2:43
yes, of course, thank you all –  user42912 Sep 28 '12 at 2:56
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