I'm asked to prove that every totally disconnected topological space satisfies $T_1$ seperation axiom. my proof is this:
fix $x\in X$. now, for each $y\in X\setminus\{x\} $ we know that $\left\{ x,y\right\}$ is not connected, therefore there are non-empty disjoint open sets in $X$, say $U,V_{y}$ such that $\left(U\cap\{ x,y\} \right)\cup\left(V_y \cap \{x,y\} \right)=\{x,y\} $. WLOG we can assume that $x\in U, y\in V_{y}$ and we get
$$X\setminus\{x\} =\bigcup_{y\in X\setminus\{x\} } V_y$$
hence every singleton is closed and $X$ is $T_1$
I've seen different proofs and I would like to know if this one is correct.
Best Answer
$U$ and $V_y$ are not necessarily disjoint in $X$. You only have $U \cap \{x,y \} = \{x \}$ and $V \cap \{x,y \} = \{y \}$.
The rest of the proof is fine.