Partition Topology is not T0

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Theorem

Let $S$ be a set and let $\mathcal P$ be a partition on $S$ which is not the (trivial) partition of singletons.

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

Then $T$ is not a $T_0$ (Kolmogorov) space.


Proof

As $\mathcal P$ is not the partition of singletons, there exists some $H \in \mathcal P$ such that $a, b \in H: a \ne b$.

Any union of sets from $\mathcal P$ which includes $H$ will therefore contain both $a$ and $b$.

Therefore, any element of $\tau$ containing $a$ will also contain $b$, and similarly, any element of $\tau$ containing $b$ will also contain $a$.

So there is no open set in $T$ containing $a$ and not $b$, or $b$ and not $a$.

Hence the result, by definition of $T_0$ (Kolmogorov) space.

$\blacksquare$


Note the reason for the exception in the statement as given.

If $\mathcal P$ is the partition of singletons:

$\mathcal P = \left\{{\left\{{x}\right\}: x \in S}\right\}$

then $T$ as constructed is the discrete topology on $S$.

From Discrete Space satisfies all Separation Properties, in that case $T$ does satisfy the $T_0$ property.


Sources