Added extremely disconnected spaces.
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Bokuan Li
2026-06-30 19:32:42 -04:00
parent 36f5b22042
commit 1f9e0bea78
3 changed files with 57 additions and 1 deletions

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@@ -60,7 +60,7 @@
is open. On the other hand, let $x \in X$, then $f(x) > \alpha$ if and only if there exists $q \in (\alpha, f(x)) \cap \rational$ such that $x \not\in U_q$. By (b) of (1), this is equivalent to the existence of $p \in (\alpha, f(x)) \cap \rational$ such that $x \not\in \ol{U_p}$. In which case,
\[
f^{-1}((b, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
f^{-1}((\alpha, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
\]
is open.