Partition Topology is T3 1/2

Theorem
Let $S$ be a set and let $\mathcal P$ be a partition on $S$.

Let $T = \left({S, \tau}\right)$ be the partition space whose basis is $\mathcal P$.

Then $T$ is a $T_{3 \frac 1 2}$ space.

Proof
Let $F \subseteq S$ be closed, and let $x \in S$, $x \notin F$.

Denote by $S \setminus F$ the relative complement of $F$ in $S$.

Define a mapping $f: S \to \left[{0 \,.\,.\, 1}\right]$ as:


 * $f \left({s}\right) := \begin{cases}

1 & \text{ if $s \in F$}\\ 0 & \text{ if $s \in S \setminus F$} \end{cases}$

Then $f$ is identically $1$ on $F$, and identically $0$ on $\left\{{x}\right\}$.

Now if $f$ is continuous, it will be a Urysohn function for $F$ and $\left\{{y}\right\}$, and $T$ will be a $T_{3 \frac 1 2}$ space.

Now for any $V \subseteq \left[{0 \,.\,.\, 1}\right]$, we have:


 * $f^{-1} \left({V}\right) = \begin{cases}

\varnothing  & \text{ if $0,1 \notin V$}\\ F            & \text{ if $0 \notin V$ and $1 \in V$}\\ S \setminus F & \text{ if $0 \in V$ and $1 \notin V$}\\ S            & \text{ if $0,1 \in V$} \end{cases}$

By definition of $\tau$, $F$ is open in $T$.

By Open Set in Partition Topology is also Closed, $F$ is also closed, and so $S \setminus F$ is open in $T$.

Thus, the preimage of any subset $V$ of $\left[{0 \,.\,.\, 1}\right]$ is open in $T$.

In particular, this holds for the open sets of $\left[{0 \,.\,.\, 1}\right]$.

It follows that $f$ is a continuous mapping, and so a Urysohn function.

Hence $T$ is $T_{3 \frac 1 2}$ space.