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Work on Countable bouquet of circles (S139) #1601

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Work on Countable bouquet of circles (S139) #1601

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10 changes: 0 additions & 10 deletions spaces/S000139/properties/P000064.md
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8 changes: 8 additions & 0 deletions spaces/S000139/properties/P000206.md
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---
space: S000139
property: P000206
value: true
---

If Player I picks $x_n= \infty$ in every turn, clearly $\infty \in \bigcap \{U_n:n<\omega\}$.
So suppose for some $n$, $x_n\neq \infty$, then Player II can choose $V_n$ to be homeomorphic to {S25} and the assertion follows since {S25|P206}.
7 changes: 7 additions & 0 deletions spaces/S000139/properties/P000214.md
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@felixpernegger felixpernegger Jan 17, 2026

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In my opinion it is easy enough to not argue why this doesn't converge, but if needed I can give a more precise argument (is a bit messy though)

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space: S000139
property: P000214
value: false
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For $m \in \mathbb{N}$, let $S_m := \{m + \frac{1}{2}, m + \frac{1}{3}, \dots\}$. Clearly $S_m$ converges to $\infty$, but picking any countable infinite $S$ with $S \cap S_m \neq \emptyset$ for infinitely many $m$ does not.