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In the definition of metric space, only one thing strikes me as unnatural: the requirement that $d(x,y) = 0$ implies $x = y$. As a programmer, I don't find it uncommon to deal with equivalence relations (or, more generally, preorders) whose equivalence classes don't have computable canonical representatives. If I wanted to make a metric space out of these, it would be very natural to require that $d(x,y) = 0$ whenever $x$ and $y$ belong in the same equivalence class.

So let's say I change the definition of a metric space $(M,d)$, so that only the following axioms are required:

  • $d(x,x) = 0$
  • $d(x,y) = d(y,x) \ge 0$
  • $d(x,y) + d(y,z) \ge d(x,z)$

What I want to understand is the effect of this change on the definition of “metrizable topological space”. Clearly, some non-metrizable spaces under the old definition become metrizable under the new one, e.g., spaces with the indiscrete topology. I'd like to know...

  1. Do any spaces remain non-metrizable?
  2. Do any interesting / useful / naturally arising spaces remain non-metrizable?
  3. Could you provide examples?
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    $\begingroup$ For the record: the keyword here is pseudometric. $\endgroup$ – Clement C. Feb 7 '16 at 14:33
  • $\begingroup$ @ClementC. Ah, thank you! $\endgroup$ – pyon Feb 7 '16 at 14:35
  • $\begingroup$ And to answer your first question, though without an explicit example: there seem to be spaces non-pseudometrizable: "There are uniformizable spaces which are not (pseudo)metrizable." $\endgroup$ – Clement C. Feb 7 '16 at 14:35
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The standard term for such a space is pseudometric space. If you take the Kolmogorov quotient by identifying $x$ and $y$ if $d(x,y)=0$, you get a metric space. If the pseudometric $d$ induces a $T_0$ topology on $X$, then it’s already a metric. Thus, any $T_0$ space that is not metrizable is an example of a non-pseudometrizable space. A very simple example is the Sierpiński space: the underlying set is $\{0,1\}$, and the topology is $\big\{\varnothing,\{1\},\{0,1\}\big\}$.

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  • $\begingroup$ Note that $0 \in \overline{\{1\}}$, so if $d$ were a pseudometric for $X$, we'd have $d(0,1) = 0$, but not $1 \in \overline{\{0\}}$, which contradicts that. Hence no pseudometric. $\endgroup$ – Henno Brandsma Feb 7 '16 at 15:59

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