# Following a proof of “every semimetric space is normal and completely regular”

Is there a flaw in the reasoning below? I was reading Theorem 4.3.3 in Topology for Analysis (Wilansky) that's given in the question. The proof uses that for any two disjoint closed sets, $A,B$ in a semimetric space $X$ with semimetric $p$, the following function is defined:

$$f(x)=\frac{p(x,A)}{p(x,A)+p(x,B)}$$

f is supposed to be a continuous function; can you point out the flaw in the counterexample?

Let $p:\mathbb{R}^2 \rightarrow \mathbb{R}$ where $p((x_1,y_1),(x_2,y_2))=|x_1-x_2|$

So p is a semimetric (see below). Consider the two disjoint closed sets in $\mathbb{R}^2$:

$A=[0,2]\times[0,1], B=[0,2]\times[2,3]$

Then $(0,2)\in B$ and $(0,0) \in A$ would imply

$0 \leq d((0,2),A)\leq d((0,2),(0,0))=0$

and clearly

$d((0,2),B)=0$

which contradicts f continuous.

To see that $p$ is a semimetric:

1. $p(a,a)=0: p((x,y),(x,y))=|x-x|=0$

2. $p(a,b)=p(b,a)\geq0: p((x_1,y_1),(x_2,y_2))=|x_1-x_2|=|x_2-x_1|=p((x_2,y_2),(x_1,y_1))\geq 0$

3. $p(a,c)\leq p(a,b)+p(b,c): p((x_1,y_1),(x_3,y_3))=|x_1-x_3|\leq |x_1-x_2|+|x_2-x_3| = p((x_1,y_1),(x_2,y_2))+p((x_2,y_2),(x_3,y_3))$

• (Note that what you’re calling a semimetric is usually called a pseudometric by topologists.) The topology of the pseudometric space $\langle\Bbb R^2,p\rangle$ is the one generated by the base of open $p$-balls, which is not the usual product topology. – Brian M. Scott Jan 3 '17 at 22:52

Your sets are not closed with respect to your semimetric $p$! They are closed with respect to the standard metric on $\mathbf R^2$, but this does not matter here. You need $p$-closed sets for this to work, which are sets of the form $A \times \mathbf R$, where $A$ is closed in $\mathbf R$ with the standard metric.