Solving a problem I came to an apparent contradiction. The problem is the following:
Let $X$ be a set and $p\in X$. We consider the following topology on $X$
$\tau_p = \{ A \subseteq X: p \notin A \} \cup \{ B \subseteq X : X \setminus B$ is finite$ \} $
Is $(X,\tau_p)$ Fréchet or Hausdorff?
I know that a set $X$ is Fréchet if and only if every point set is closed. However, if $X$ has more than one element, say $a$, then $\{a\}$ is open, since $p \notin \{a\}$. But I've successfully shown that $X$ is Hausdorff, which should imply that $X$ is Fréchet.
I'm so confused about this.
In case of finite set $X$, the cofinite topology and the discrete topology coincidies and hence each singleton is closed and open.
In case $X$ is infinite.
Then $X\setminus\{p\}$ is open hence $\{p\}$ is closed.
and for $p\neq a$.
$X\setminus (X\setminus\{a\})=\{a\}$ is finite and hence $X\setminus\{a\}$ is open and hence $\{a\}$ is closed.
Hence $X$ is Frechet.
Now for $x,y\in X$ and $x\neq y$ .
If one of $x$ or $y$ is $p$ say $x=p$.
Then $\{y\}$ is open And $X\setminus \{y\}$ is an open set such that $p\in X\setminus \{y\}$ and both are disjoint.
If $x\neq p$ and $y\neq p$ .
Then $\{x\}$ and $\{y\}$ are two open disjoint sets.
Then $X$ is Hausdorff.