Is there an alternative proof to Urysohn's lemma, that makes use of $d(x, A)$?

Urysohn's lemma is: given a normal topological space $X$, for any disjoint closed sets $A$, $B$, there exists a separation by continuous function $f$. I.e. there exists $f : X \to [a,b]$ with $a,b \in \Bbb{R}$ such that $f$ is continuous and $f(A) = a$ while $f(B) = b$, for all $a,b \in \Bbb{R}$.

My question is, since $d(x,A), d(x,B)$ are continuous functions from $X$ into $\Bbb{R}$, can we somehow construct the required Urysohn function from this?


  • 2
    $\begingroup$ $$\frac{d(x,A)}{d(x,A) + d(x,B)}$$ $\endgroup$ – Daniel Fischer Oct 23 '13 at 22:09
  • $\begingroup$ $a + (b-a)\cdot f(x)$, that's a generally useful transformation. $\endgroup$ – Daniel Fischer Oct 23 '13 at 22:13
  • $\begingroup$ Thanks that makes sense. It's to $[0,1]$ so we can map $[0,1]$ to any $[a,b]$ continuously $\endgroup$ – Shine On You Crazy Diamond Oct 23 '13 at 22:13
  • $\begingroup$ I guess this only works if $X$ is metrizeable. $\endgroup$ – Shine On You Crazy Diamond Oct 23 '13 at 22:16
  • 3
    $\begingroup$ Of course. If we have no metric, we don't have $d(x,A)$. $\endgroup$ – Daniel Fischer Oct 23 '13 at 22:18

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