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Theorem Suggestion: perfectly normal LOTS are first countable #1041
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The observation with the second link in your post is curious however. Any ideas why this search works like this @StevenClontz ? |
I was not aware that I can accidentally access old versions of the base. |
@pzjp On the pi-base website, you can click the circle of arrows at the bottom of your screen to reset pi-base (I believe without changing the branch). Or you can go to https://topology.pi-base.org/dev and click Reset, which will reset pi-base and set the branch to main. |
I've did this when the possibility of having outdated base was pointed out ;) I used to think that the date of migration refers to a global update I'm not able to interfere with. |
Reopened pi-base/web#166 to find a way to automatically resync when data is over a day old. |
It's buggy lol. 🤷♂️ I'm expecting to sync up with @jamesdabbs in the coming weeks to tackle a few things that he's more knowledgable about; I'll open an issue for this. |
Thanks for the contribution @pzjp - I'm closing as this seems to be deducable with an updated database. |
@pzjp FYI, for a probable explanation of the "buggy behavior": |
Theorem Suggestion
If a space is:
then it is First countable P28.
Rationale
This theorem would demonstrate that no spaces satisfy the following search:
https://topology.pi-base.org/spaces?q=LOTS%2BPerfectly+normal%2B%7EFirst+Countable
Observation: when I put "?" instead of "~" then a weird answer appears:
https://topology.pi-base.org/spaces?q=LOTS%2BPerfectly+normal%2B%3FFirst+Countable
Proof
Elementary:
By perfect normality, for$x\in X$ we have $\{x\}=\bigcap_{n=1}^\infty U_n$ for some descending sequence of open sets $U_n$ . By definition of the order topology, we can replace $U_n$ by an open interval $(a_n,b_n)\subset U_n$ , where $-\infty\leq a_n\leq a_{n+1}<x<b_{n+1}\leq b_n\leq +\infty$ (we introduce $\pm\infty$ to cover neighborhoods of minimal/maximal element in $X$ ).$(a,b)\ni x$ there has to exist $m,n$ such that $a\leq a_n < x < b_m \leq b$ . Otherwise either $[a,x]$ or $[x,b]$ would be contained in the intersection of $U_n$ 's. Then $x\in(a_{n'},b_{n'})\subset(a,b)$ for $n'=\max(m,n)$ . Hence we proven that $\{(a_n,b_n)\colon n\geq 1\}$ is a countable base of neighborhoods at $x$ .
For any
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