In the famous book Counterexamples in topology by Steen and Seebach, is proved that Arens space is not a Urysohn space (in that book, a Urysohn space is the space in which any two distinct points can be separated by a continuous function). The proof from the book is as follows:
We can show that X is not Urysohn by considering a function $f:X \to [0,1]$ such that $f(0,0)=0$ and $f(1,0)=1$. If $f$ is continuous we note that the inverse images of the open sets $[0,\frac{1}{4})$ and $(\frac{3}{4},1]$ of $[0,1]$ must be open and hence contain $U_n(0,0)$ and $U_m(1,0)$ for some $m$ and $n$, respectively. Then, if $r\sqrt{2}<\min(\frac{1}{n},\frac{1}{m})$, $f(\frac{1}{2}r\sqrt{2})$ is not in both $[0,\frac{1}{4})$ and $(\frac{3}{4},1]$, so suppose it is not in $[0,\frac{1}{4})$...
before this everything is clear for me. But I can't understand the next sentence:
... then, there exists about $f(\frac12 r\sqrt{2})$ an open interval $U$ such that $\overline{U}$ and $[0,\frac14)$ are disjoint. But then the inverse images of $\overline{U}$ and $\overline{[0,\frac14)}$ under $f$ would be disjoint closed sets containing...... etc.
I can't understand, why does such a $U$ exist? There is no chance that $f(\frac12 r\sqrt{2})=\frac14$.
Please, help me!
P.S. The definition of Arens square: Wikipedia link