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Aug 1, 2022 at 8:53 history became hot network question
Aug 1, 2022 at 7:07 vote accept erz
Aug 1, 2022 at 4:50 answer added Joseph Van Name timeline score: 16
Jul 31, 2022 at 23:53 comment added David Handelman I think the condition on $X $ is that all countable suprema of projections exist in $C(X)$; this is the same as the closure of a countable union of clopen sets is clopen. Monotone $\sigma-$complete, I think is the term.
Jul 31, 2022 at 23:40 answer added Nik Weaver timeline score: 9
Jul 31, 2022 at 23:38 comment added LSpice @‍erz, as long as your space is infinite, it has an injective sequence, which has a limit point by compactness. Maybe it's not a limit point of a subsequence, but doesn't @NikWeaver's argument still apply? (I'm not sure where connectedness, which @‍NikWeaver doesn't seem to use, comes into it.)
Jul 31, 2022 at 23:33 comment added erz @NikWeaver thank you for your comment, but the whole point is that spaces like this are very non-sequential and also very disconnected.
Jul 31, 2022 at 23:32 comment added Nik Weaver @LSpice yeah, good point.
Jul 31, 2022 at 22:01 comment added LSpice @NikWeaver, that seems like an answer; but, if you post it as such, then you might consider avoiding $h$, which is used with another meaning in the body of the question.
Jul 31, 2022 at 20:50 comment added Nik Weaver If $x_n \to x$ (all distinct), define $g(x_n) = 1/n$, $g(x) =0$ and set $h(x_n) = 1/n$ or $0$ depending on whether $n$ is even or odd, and $h(x) =0$. Extend $g$ and $h$ to positive continuous functions on $X$ by Tietze, then set $f = g \wedge h$.
Jul 31, 2022 at 19:27 comment added Nik Weaver Seems to me that if $X$ contains a nonconstant convergent sequence it can't have this property.
Jul 31, 2022 at 19:03 history asked erz CC BY-SA 4.0