Franklin (1964) wrote in his paper the definition of sequentially closed and sequentially open.
"Let $X$ be a topological space. A subset $U$ of X is sequentially open if and only if each sequence in $X$ converging to a point in $U$ is eventually in $U$. A subset $F$ of $X$ is sequentially closed if and only if no sequence in $F$ converges to a point not in $F$."
From this definition, we get
- Every closed set is sequentially closed.
- $A\subseteq X$ is sequentially open iff $X\setminus A$ is sequentially closed.
I have one question about it. If $X$ is a topological space in which every singleton is sequentially closed, is it $T_1$?
I have tried to prove it indirectly but I have never found the contradiction.
" First, I assume that X is non T1. There is $x\in X$ such that {x} is not closed. It doesn't contradict with the fact $\{x\}$ is sequentially closed."
Can anybody help me to figure out the counterexample? Thank you in advance.
Best Answer
Let $y$ be another point of the closure $C$ of $\{x\}$.
Then, any open neighborhood $U$ of $y$ contains $x$ as well, because otherwise $x\in X\setminus U\ \implies \ C\subseteq X\setminus U\ \implies\ y\notin U$ would follow.
This means that $y$ is also a limit point of the constant sequence $x$.
So the closure of a singleton coincides with its sequential closure.