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Suppose

  1. $D$ is a collection of metrics on a set $X$.
  2. $\mathcal T$ is the (smallest) topology generated by metrics in $D$ .
  3. $F\subseteq X$ is closed in $(X,\mathcal T)$.
  4. $p\in X\setminus F$.

Is there any $d\in D$ so that the (so-called) function

$$h:X\to [0,1]$$ $$h(x)=\frac{d(p,x)}{d(p,x)+d(F,x)}$$

is well-defined?

Willie Wong
  • 73,139

1 Answers1

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If $x \neq p$ then $d(p,x)+d(F,x) \geq d(p,x) \neq 0$ for all $d \in D$, so it is equivalent to show that $d(p,F) \neq 0$ for some $d \in D$.

A basis for the topology $\mathcal{T}$ is $\left\{ \bigcap\limits_{d \in M} B_d(x_d,\epsilon) \mid M \subset D \ \text{finite}, x_d \in X, \epsilon >0 \right\}$. So, if $d(F,p)=0$ for every $d \in D$, every neighborhood of $p$ intersects $F$, hence $p \in F$ since $F$ is closed for $\mathcal{T}$.

Seirios
  • 33,157
  • thanks. It seems your proof doesn't work for pseudometrics because you used $d(p,x)\ne 0$. –  Mar 11 '13 at 16:49
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    I tried to modify your proof so that it works for pseudometrics too: If $d(p,x) \neq 0$ then $d(p,x)+d(F,x) \geq d(p,x) > 0$ for all $d \in D$, so it is equivalent to show that for each $q\in {x\mid d(x,p)=0}$, $d(q,F) \neq 0$ for some $d \in D$.

    A basis for the topology $\mathcal{T}$ is $\left{ \bigcap\limits_{d \in M} B_d(x_d,\epsilon) \mid M \subset D \ \text{finite}, x_d \in X, \epsilon >0 \right}$. So, if $d(F,q)=0$ for every $d \in D$, every neighborhood of $p$ which is also a neighborhood of $q$ intersects $F$, hence $p \in F$ since $F$ is closed for $\mathcal{T}$. A contradiction.

     –  Mar 11 '13 at 17:02