Either-Or Topology is Non-Meager
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Theorem
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is non-meager.
Proof 1
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.
$\blacksquare$
Proof 2
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$.
The result follows directly from Space with Open Point is Non-Meager.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $3$