2
$\begingroup$

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)) $

$\endgroup$
  • 1
    $\begingroup$ (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. $\endgroup$ – Brian M. Scott Jan 3 '17 at 22:52
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.