Partition Topology is T3 1/2

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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.


Corollary

$T$ is a $T_3$ 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:

$f^{-1} \sqbrk V = \begin{cases} \O & : \text{ if } 0, 1 \notin V \\ F & : \text{ if } 0 \notin V \text { and } 1 \in V \\ S \setminus F & : \text{ if } 0 \in V \text { and } 1 \notin V \\ S & : \text{ if } 0, 1 \in V \end{cases}$


Fake Proof suggests: The validity of the material on this page is questionable.
In particular: $F$ is defined to be closed. Why is it open by definition? --Fake Proof (talk contribs) 18:02, 13 July 2022 (UTC)
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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.

$\blacksquare$


Sources

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