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There is a theorem stating that every metric space embeds isometrically into $\ell _{\infty}$.

My question: is there a generalized result for pseudo metric spaces embedding isometrically into semi-normed spaces? Is/would this result be interesting/useful?

Thank you very much for taking the time to answer my question.

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  • $\begingroup$ It is definitely not the case that every metric space embeds into $\ell^\infty$ by cardinality reasons, there must be some size assumption on the spaces $\endgroup$
    – Alessandro Codenotti
    Commented Oct 16 at 16:44
  • $\begingroup$ @AlessandroCodenotti certainly OP means $\ell^\infty$ of some set (possibly not countable). $\endgroup$
    – YCor
    Commented Oct 16 at 16:48
  • $\begingroup$ Would you say what is meant by "pseudo"? Allows zero distance? negative distance? If it just relaxes to allow distance zero, I guess one can get results as trivial corollaries of the metric case. $\endgroup$
    – YCor
    Commented Oct 16 at 16:50
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    $\begingroup$ @AlessandroCodenotti I’m pretty sure YCor meant that the index set $Y$ of $\ell_\infty(Y)$ can depend on $X$, so the second one. $\endgroup$
    – David Gao
    Commented Oct 16 at 16:56
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    $\begingroup$ @DavidGao Yes, exactly. Fix $x_0\in X$, and $Y$ a dense subset of $X$, and take the map $x\mapsto f_x\in\ell^\infty(Y)$, where $f_x(y)=d(y,x)-d(y,x_0)$. $\endgroup$
    – YCor
    Commented Oct 16 at 17:03

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Yes. If $(X, d)$ is a pseudometric space, then we may may define an equivalence relation $\sim$ on $X$ by $x \sim y$ iff $d(x, y) = 0$. Then $(X/{\sim}, d’)$ with $d’([x], [y]) = d(x, y)$ is a metric space, so there is some isometric embedding $\phi: X/{\sim} \to \ell_\infty(I)$. Now, let $\kappa = \sup_{x \in X} |[x]|$ and pick a vector space $V$ of cardinality at least $\kappa$ (say, $V = \mathbb{K}[\kappa]$). For each equivalence class $[x]$, fix an injection $\psi_{[x]}: [x] \to V$. Now, equip $\ell_\infty(I) \times V$ with the seminorm:

$$p(f, v) = \|f\|_{\ell_\infty(I)}$$

Then $\pi: X \to (\ell_\infty(I) \times V, p)$ defined by,

$$\pi(x) = (\phi([x]), \psi_{[x]}(x))$$

is an isometric embedding. (I don’t see how such a result is in any way interesting or useful though.)

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