# every single point is closed?

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?

• 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
• how we define a mathematical point? – sudhakaran gopalan Dec 25 '14 at 22:29

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 $$x\in X$$. Then 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 $$U_n=\{k\in\Bbb N:k and let $$\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$$.

• 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

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.

• 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
• @MustafaGokhanBenli I am using the definition of limit points in Rudin Principles of Mathematical Analysis. Theorem 2.20 and related corollary give: Theorem: If p is a limit point of a set E, then every neighborhood of p contains infinitely many points of E. Corollary: A finite point set has no limit points. So how is 3 a limit point of {2}. Could you please clarify? – texmex Aug 7 at 8:18