Either-Or Topology is Non-Meager/Proof 1

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Let $T = \struct {S, \tau}$ be the either-or space.

Then $T$ is non-meager.


From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$.

So every non-empty subset of $T$ (apart from $\left\{{0}\right\})$ contains at least one open set of $T$.

So no subset of $T$ is nowhere dense in $T$.

So $T$ is not a countable union of subsets of $S$ which are nowhere dense in $T$.

Hence the result by definition of non-meager.