# In $T_1$ space, all singleton sets are closed?

The definition of $$T_1$$-Space is:

A topological space $$X$$ is said to be $$T_1$$ if for each pair of distinct points $$a,b,$$ $$\exists$$ open sets $$U,V$$ s.t $$a\in U, b\notin U, a\notin V, b\in V$$.

What I'm confused about is in a $$T_1$$ space, all singleton subsets of $$X$$ are closed.

Let $$t,v \in X$$.

Then I think the singleton sets $$\{t\}$$ ,$$\{v\}$$ satisfy the definition of $$T_1$$ in $$U$$ and $$V$$ what I wrote above.

(i.e $$t \in\{t\}$$, $$v\notin \{t\}$$, $$t\notin\{v\}$$, $$v \in\{v\}$$.)

I learned the theorem showing this result and I can understand the proof of it, but I'm still confused as to why this is not a counterexample.

• Are you asking for a proof that in a $T_1$ space singletons are closed?
– R_D
Jun 18, 2016 at 15:22
• No... I'm asking why my guess is wrong Jun 18, 2016 at 15:23
• {t} and {v} are closed, NOT open. Why do you want to write $t\in \{t\}, v\notin \{t\}$...?
– R_D
Jun 18, 2016 at 15:25
• @Rise 'closed' in topological space mean that 'it is not in topology of X' right? So I did like it... Jun 18, 2016 at 15:28
• No. You can define the topology of $X$ using either open or closed sets. If you define the topology with open sets then a closed set is a complement of a member of the topology. It is not whatever is outside the topology.
– R_D
Jun 18, 2016 at 15:32

## 1 Answer

Providing both sides.

If $$X$$ is $$T_1$$ then all singletons are closed.

Proof: Let $$x\in X$$. For all $$y\in\{x\}^{\complement}$$ there is an open set $$U_y$$ with $$y\in U_y$$ and $$x\notin U_y$$. Then $$U=\bigcup_{y\in\{x\}^{\complement}} U_y$$ is open and is the complement of $$\{x\}$$. That means exactly that $$\{x\}$$ is closed.

If in $$X$$ every singleton is closed then $$X$$ is $$T_1$$.

Proof: Let $$x,y\in X$$ with $$x\neq y$$. Then $$\{x\}^{\complement}$$ is an open set with $$y\in\{x\}^{\complement}$$ and $$x\notin\{x\}^{\complement}$$.

The given definition of $$T_1$$ is a bit weird. It is enough to demand that for each pair $$a,b$$ of distinct there is an open sets $$U$$ with $$a\in U$$ and $$b\notin U$$. This implies immediately that there is also an open set $$V$$ with $$a\notin V$$ and $$b\in V$$.