In step 4. of Munkres' proof of Urysohn Lemma, how does he prove the continuity of the function? I am reading Munkres' Topology and in the proof of the Urysohn Lemma I get everything except the last part in showing that $f$ is a continuous function. I have the second edition of the book and am on page 210.
In step 4 of the proof of the Urysohn Lemma he says:
$$x \in \bar{U}_r \implies f(x) \leq r$$ and 
$$x \notin U_r \implies f(x) \geq r.$$
Why is this the case? If $x \in \bar{U}_r$ then $x \in U_s$ for every $s >r$, then $\Bbb{Q}(x)$ contains all rationals greater than $r$, this I get. But then how is 
$$f(x) = \inf \Bbb{Q}(x) \leq r?$$
Please help/ A;sp, if $x \notin U_r$ then why is $x$ not in $U_s$ for any $s < r$? How do I conclude that $f(x) = \inf \Bbb{Q}(x) \geq r$ in this case?
Please help me.
 A: $\newcommand{\cl}{\operatorname{cl}}$I have only the first edition, but he does not appear to have changed the argument much in the second edition. First let me recapitulate the construction. For each $r\in\Bbb Q$ you have an open set $U_r$. These open sets have the following properties:


*

*$A\subseteq U_0$;  

*$U_1=X\setminus B$;  

*$\cl U_p\subseteq U_q$ whenever $p<q$;  

*$U_r=\varnothing$ if $r<0$; and  

*$U_r=X$ if $r>1$.


For $x\in X$ you have $\Bbb Q(x)=\{r\in\Bbb Q:x\in U_r\}$, and $f$ has been defined by $f(x)=\inf\Bbb Q(x)$. We want to show first that if $x\in\cl U_r$, then $f(x)\le r$.

Suppose that $x\in\cl U_r$. If $s\in\Bbb Q$ and $s>r$, then by construction $\cl U_r\subseteq U_s$, so $x\in U_s$, and $s\in\Bbb Q(x)$. In other words, $\Bbb Q(x)\supseteq\{s\in\Bbb Q:s>r\}$. Thus, any lower bound for $\Bbb Q(x)$ is also a lower bound for $\{s\in\Bbb Q:s>r\}$. In particular, $\inf\Bbb Q(x)$, the greatest lower bound for $\Bbb Q(x)$, is a lower bound for $\{s\in\Bbb Q:s>r\}$: $\inf\Bbb Q(x)\le s$ for all rational $s>r$. Clearly $r$ is the greatest bound for $\{s\in\Bbb Q:s>r\}$, so $\inf\Bbb Q(x)\le r$. And $f(x)=\inf\Bbb Q(x)$, so $f(x)\le r$. $\dashv$

Now we want to show that if $x\notin\cl U_r$, then $f(x)\ge r$.

For each $s<r$ we know that $U_s\subseteq U_r$, so if $x\notin\cl U_r$, then $x\notin U_s$ for any $s<r$. In other words, if $s<r$, then $s\notin\Bbb Q(x)$. Take the contrapositive: if $s\in\Bbb Q(x)$, then $s\ge r$. That is, every member of $\Bbb Q(x)$ is at least as big as $r$, and $r$ is therefore a lower bound for $\Bbb Q(x)$. Clearly, then, $r\le\inf\Bbb Q(x)=f(x)$. $\dashv$

