Metric space is perfectly normal?

Question

This post: Is a metric space perfectly normal? answers a similar question, but the definition I have for "perfect normality" is that every two closed, disjoint sets $A$ and $B$ must be separated by a continuous function, i.e. there is a continuous $f:X\to [0,1]$ such that $A\subseteq f^{-1}(1)$ and $B\subseteq f^{-1}(0)$. This is fairly simple if $d(A,B)=\inf \{d(a,b):a\in A\land b\in B\}\neq 0$, but there are sets for which this is not the case. That is, if this value is non-zero, then I can use the function

$$ x\mapsto \frac{\min\{d(A,B),d(x,B)\}}{d(A,B)} $$ for the proof. This doesn't work if the distance between A and B is zero, however, as in the case of a non-connected metric space. I imagine there's a simple fix, but my brain's not seeing it, unfortunately. Any assistance?

Thank you!

Your definition of perfect normality is incorrect: it should require that $A=f^{-1}[{1}]$ and $B=f^{-1}{{0}]$. — Brian M. Scott, Jul 09 '16 at 21:23
Isn't perfectly normal equivalent to normal? I mean, we have Urysohn's lemma. — Crostul, Jul 09 '16 at 23:01

score 2 · Answer 1 · answered Jul 09 '16 at 21:50

One can use $\;\;\; x \: \mapsto \: \dfrac{d(x,\hspace{-0.02 in}A)}{d(x,\hspace{-0.02 in}A)+d(x,\hspace{-0.03 in}B)} \;\;\;$ instead.

(Note that, according to Brian Scott's comment and wikipedia, the $\subseteq$s should be replaced with $=$s.
The above function also achieves that stronger goal.)

Metric space is perfectly normal?

1 Answers1

Linked