Partition Topology is T3 1/2

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

Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.

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

Proof
Let $F \subseteq S$ be closed.

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

Let $x \in S \setminus F$.

Define a mapping $f: S \to \closedint 0 1$ as:


 * $\map f s := \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 $\set x$.

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

Now for any $V \subseteq \closedint 0 1$, we have:


 * $f^{-1} \sqbrk V = \begin{cases}

\O           & : \text{ if } 0, 1 \notin V \\ F            & : \text{ if } 0 \notin V \text { and } 1 \in V \\ S \setminus F & : \text{ if } 0 \in V \text { 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 $\closedint 0 1$ is open in $T$.

In particular, this holds for the open sets of $\closedint 0 1$.

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

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