Something about the proof of Urysohn's lemma

Question

Let $X$ be a normal space and $F,G\subseteq X$ closed sets such that $F\cap G=\emptyset$.

In the proof of Urysohn's lemma, we proved that if $D=\{q_n:n=0,1,...\}=\mathbb{Q}\cap [0,1]$ with $q_0=0$ and $q_1=1$, then there exists a family of open sets $\mathcal{U}=\{U_{q_n}:n=0,1,...\}$ such that $F\subseteq U_0$, $U_1=X\setminus G$ and if $r<s$ with $r,s\in D$ then $Cl(U_r)\subseteq U_s$.

Now, we define $f(x)= \inf \left\{{r\in D:x\in U_r}\right\} $ if $x\in X\setminus G$ and $f(x)=1$ if $x\in G$.

I don't understand why if $x\in X\setminus G$ and $b\in (0,1)$, then $f(x)>b$ if and only if there exists $r\in D$ such that $r>b$ and $x\notin U_r$.

Can someone explain this to me?

Thanks!

Edit: I meant to say $q_0=0$. I already edited it.

Does this answer your question? In step 4. of Munkre's proof of Urysohn Lemma, how does he prove the continuity of the function? — tryst with freedom, Aug 22 '22 at 16:01

score 0 · Accepted Answer · answered Nov 12 '13 at 02:31

If $f(x)>b$, there is $r\in D$ such that $f(x)> r > b$. By definition of $f$, $x\notin U_r$. On the other hand, if such an $r$ exists and $x\in U_p$ for some $p\in D$, then $p>r$ (or $x\in U_r$). Then $f(x)\geq r >b$.

Something about the proof of Urysohn's lemma

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