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Let $(X, \tau)$ be a metrizable topological space. Any metric $d$ on $X$ generates a topology $\tau_d$ on it. Call $d$ admissible if $\tau = \tau_d$.

Let $\mathcal{D}$ be the set of admissible metrics for $(X, \tau)$, and denote $B_d(x, r) = \{y : d(x, y) < r\}$. Consider the set $$\mathcal{U} = \{B_d(x, r) : d\in\mathcal{D}, x\in X, r> 0\}.$$

In other words, $\mathcal{U}$ is the family of those open sets, which are open balls for some admissible metric on $X$.

Can we characterize elements of $\mathcal{U}$ topologically (i.e. not involving the concept of a metric)?

This problem didn't come up in anything I'm working with, I'm asking it out of curiosity.

Jakobian
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1 Answers1

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If $U\subseteq X$ is any open set and $x\in U$, then there exists an admissible metric $d$ such that $U=B_d(x,1)$ (so in particular, $\mathcal{U}=\tau\setminus\{\emptyset\}$). To prove this, first fix an admissible metric $d_0$ such that $d_0(a,b)<1$ for all $a,b$. Now take a continuous function $f:X\to[0,1]$ such that $f(x)=1$ and $f^{-1}(\{0\})=X\setminus U$ (for instance, $f(y)=\min(d_0(y,X\setminus U),d_0(x,X\setminus U))/d_0(x,X\setminus U)$). Now define a metric $d$ by $d(a,b)=\max(d_0(a,b),|f(a)-f(b)|)$. Since $d_0<1$ everywhere, it is easy to see that $B_d(x,1)=U$. Also, since $d_0$ is admissible and $f$ is continuous, it is easy to see $d$ is admissible.

Eric Wofsey
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