Particular Point Space is Non-Meager

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Theorem

Let $T = \left({S, \tau_p}\right)$ be a particular point space.


Then $T$ is non-meager.


Proof 1

Suppose $T$ were meager.

Then it would be a countable union of subsets which are nowhere dense in $T$.


Let $H \subseteq S$.

From Closure of Open Set of Particular Point Space, the closure of $H$ is $S$.

From the definition of interior, the interior of $S$ is $S$.

So the interior of the closure of $H$ is not empty.

So $T$ can not be the union of a countable set of subsets which are nowhere dense in $T$.

Hence $T$ is not meager and so by definition must be non-meager.

$\blacksquare$


Proof 2

By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.

So $\left\{{p}\right\}$ itself is open in $T$.

That is, $p$ is an open point.

The result follows from Space with Open Point is Non-Meager.

$\blacksquare$


Proof 3

By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.


Aiming for a contradiction, suppose $T$ is meager.

By definition, $T$ is meager if and only if it is a countable union of subsets of $S$ which are nowhere dense in $T$.

At least one such nowhere dense subset $U$ of $S$ must contain $p$.


By definition, $U$ is nowhere dense in $T$ if and only if:

$U^-$ contains no open set of $T$ which is non-empty

where $U^-$ denotes the closure of $U$.


By definition of particular point space, $U$ is open in $T$.

By Closure of Open Set of Particular Point Space, $U^- = S$.

But $S$ is itself open in $T$ and non-empty, and so $U$ is not nowhere dense.


From this contradiction it follows that $T$ is non-meager

$\blacksquare$