Theorem Suggestion
If a space is:
- Orderable P133
- Perfectly normal P15
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 $X$).
For any $(a,b)\ni x$ 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 $x$.
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