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.
Best Answer
Providing both sides.
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.
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 $b\in V$.