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First of all, I am "new" to proof of Urysohn's lemma.

I have a question on proof of this theorem from wiki; I am trying to understand essential part of each step.

Let $X$ be normal topological space, and $A,B$ disjoint closed subsets.

To construct a continuous function from $X$ to $[0,1]$ with $f(A)=0$ and $f(B)=1$, we construct a family of open sets $U_r$, for each rational $r=\frac{a}{2^n}$ in $[0,1]$, $a$ odd, such that

(1) $U_r$ contains $A$

(1')$U_r$ is disjoint from $B$ for all $r$;

(2) $r

Define $f$ by $f(x)=1$ if $x$ is not in any $U_r$, and define $f(x)=inf\{ r: x\in U_r\}$.

My question is following: let us forget set $B$ for a moment, and consider sets $U_r$ defined for each rational $r=\frac{a}{2^n}$ in $[0,1]$, $a$ odd, such that they satisfy (1), (2) [ignore (1')].

Simple question is, is $f$ still continuous?

Further, suppose in definition of $f$ we replace inf by sup; is still $f$ continuous? [Ignore the expected properties of $f$ on $A$ and $B$.]

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    Have you written out the proof of continuity in detail? Where do you use properties of inf? I believe only (2) is esential for continuity. – 2017-02-23
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    Have you looked at the proof writen out here: http://math.stackexchange.com/a/1106991/4280 – 2017-02-24

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