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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.

Arctic Char
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JAEMTO
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  • Are you asking for a proof that in a $T_1$ space singletons are closed? – R_D Jun 18 '16 at 15:22
  • No... I'm asking why my guess is wrong – JAEMTO Jun 18 '16 at 15:23
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    {t} and {v} are closed, NOT open. Why do you want to write $t\in {t}, v\notin {t}$...? – R_D Jun 18 '16 at 15:25
  • @Rise 'closed' in topological space mean that 'it is not in topology of X' right? So I did like it... – JAEMTO Jun 18 '16 at 15:28
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    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 '16 at 15:32
  • @Rise I see... then why {$t$}, {$v$} are closed in X? I assumed that $t,v$$\in$ X. They can't be open? Because they satisfy $U,V$ in the above, they can be open sets, I think – JAEMTO Jun 18 '16 at 15:43
  • Let us continue this discussion in chat. – JAEMTO Jun 18 '16 at 15:45
  • The definition 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 $b\in V$. – Vera Nov 13 '21 at 15:44

1 Answers1

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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 $b\in V$.

Vera
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