Modified Fort Space is T1

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Theorem

Let $T = \left({S, \tau_{a, b}}\right)$ be a modified Fort space.


Then $T$ is a $T_1$ (Fréchet) space.


Proof

Let $p, q \in S: p \ne q$.

There are two options:

$(1): \quad$ At least one of $p$ and $q$ is not in $\left\{{a, b}\right\}$
$(2): \quad$ Both $p$ and $q$ are in $\left\{{a, b}\right\}$.


$(1): \quad$ First let $p \notin \left\{{a, b}\right\}$ or $q \notin \left\{{a, b}\right\}$.

Without loss of generality, let $p \notin \left\{{a, b}\right\}$.

From Clopen Points in Modified Fort Space both $\left\{{p}\right\} \in \tau_{a, b}$ and $S \setminus \left\{{p}\right\} \in \tau_{a, b}$.

So:

$p \in \left\{{p}\right\}, q \notin \left\{{p}\right\}$
$p \notin S \setminus \left\{{p}\right\}, q \in S \setminus \left\{{p}\right\}$

Similarly for $q \notin \left\{{a, b}\right\}$.


$(2): \quad$ Next, let $p, q \in \left\{{a, b}\right\}$.

Let $V_p \subseteq S$ such that $p \in V_p, q \notin V_p$ and $V_p$ is finite.

Then $S \setminus V_p$ is open in $T$ containing $q$ but not $p$.

Similarly, let $V_q \subseteq S$ such that $q \in V_q, p \notin V_q$ and $V_q$ is finite.

Then $S \setminus V_q$ is open in $T$ containing $p$ but not $q$.


Hence it can be seen that $T$ is a $T_1$ (Fréchet) space.

$\blacksquare$


Sources