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.

  • $\begingroup$ Are you asking for a proof that in a $T_1$ space singletons are closed? $\endgroup$
    – R_D
    Jun 18, 2016 at 15:22
  • $\begingroup$ No... I'm asking why my guess is wrong $\endgroup$
    – JAEMTO
    Jun 18, 2016 at 15:23
  • 1
    $\begingroup$ {t} and {v} are closed, NOT open. Why do you want to write $t\in \{t\}, v\notin \{t\}$...? $\endgroup$
    – R_D
    Jun 18, 2016 at 15:25
  • $\begingroup$ @Rise 'closed' in topological space mean that 'it is not in topology of X' right? So I did like it... $\endgroup$
    – JAEMTO
    Jun 18, 2016 at 15:28
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    $\begingroup$ 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. $\endgroup$
    – R_D
    Jun 18, 2016 at 15:32

1 Answer 1


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 $b\in V$.


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