Uncountable Subset of Countable Complement Space Intersects Open Sets
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Theorem
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then the intersection of $H$ with any non-empty open set of $T$ is uncountable.
Proof
Let $U \in \tau$ be any non-empty open set of $T$.
Then $\relcomp S U$ is countable.
Suppose $H \cap U = \O$.
Then from Intersection with Complement is Empty iff Subset it follows that $H \subseteq \relcomp S U$ and so $H$ is countable.
So if $H$ is uncountable it is bound to have a non-empty intersection with every open set in $T$.
This needs considerable tedious hard slog to complete it. In particular: It still needs to be shown that $U \cap H$ is uncountable if and only if $H$ is uncountable. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $1$