In a comment on the first answer to this question, @Nate Eldredge stated that

"For instance, there is a metric on $\ell^{\infty}$ that makes it isometric to $[0,1]$.

How would that look like?

Edit: As pointed out by @Henno Brandsma we can define $d(x,y) = | f(x) - f(y) |$, where $f$ is a bijection $\ell^{\infty} \to [0,1]$. Precisely, my question should therefore by: What does this bijection look like?


2 Answers 2


OK, here's my attempt to give an explicit bijection $g:\ell^\infty\to[0,1]$. The intuition is that sets that admit "countable compact covers" are not so different from compact sets (along the same lines as what Henno says in their first line).

For starters, it is not so hard to describe an injection $f:K\to[0,1]$ for $K$ any bounded subset of $\ell^\infty$. If we let $k=\sup_{x\in K} \lVert x \rVert_\infty$, and write elements of $\ell^\infty$ as $x=(x_i)_{i\in\mathbb{N}}$, then we can describe $f(x)$ as follows. Let $p:\mathbb{N}^2\to\mathbb{N}$ be the usual diagonal bijection of those sets and $p^{-1}$ its inverse. Also, let $e:[0,1]\to\{0,\dots,9\}^{\mathbb{N}}$ be a bijective decimal expansion. Then we can define the $f(x)$ as the number in $[0,1]=[0,0.99\bar{9}]$ with the digit \[e(x_i/k)_j\] at the $p^{-1}(i,j)$th position of its decimal expansion.

Now, extend $f$ to a map $f'_k:\ell^\infty\to[0,1]$ by taking $$f'_k(x) = \begin{cases}f(x)&\lVert x\rVert_{\infty}\leq k\\ 0 & \text{otherwise}.\end{cases}$$ Note that these are all injective when restricted to the set of sequences with supremum less than $k$, but that the kernel is $\{0\}$ together with the rest of $\ell^\infty$. These are also surjective since, in particular, $f_k'((k,k,k,k,k\dots))=1$.

Now, consider the (countable) family $F=\{f_k'\}_{k\in\mathbb{N}}$. We'll splice all of these together to build our final bijection, which I should have called $f$ to match your notation, but at this point I think I'll go with $g$. Let $q:\mathbb{N}^3\to\mathbb{N}$ be a standard diagonal style bijection like above. Then define $g:\ell^\infty\to [0,1)$ by letting $g(x)$ be the number with the digit \[e(x_i/k)_j\] at the $q^{-1}(i,j,k)$th position in its decimal expansion. (Note that there will be a lot of 0s in this decimal expansion.)

(OK, so the range doesn't include $1=0.\bar{9}$ since all $g(x)$ are "eventually mostly 0", but feel free to compose with a bijection of $[0,1]$ and $[0,1)$.)


$\ell^\infty$ has the same cardinality as $[0,1]$ (because $\mathfrak{c}^{\aleph_0} = \mathfrak{c}$, essentially) so we can fix a bijection $f: \ell^\infty \to [0,1]$.

Then define a metric on $\ell^\infty$ by $d(x,y)=|f(x)-f(y)|$.

Then $f$ is the isometry Nate referred to.

  • $\begingroup$ @ViktorGlombik It's simple set theory that tells us such a bijection exists. I don't know of an easy explicit one; why would you want one? $\endgroup$ Apr 27, 2019 at 14:25
  • $\begingroup$ Purely out of interest. $\endgroup$ Apr 27, 2019 at 14:30
  • 2
    $\begingroup$ @ViktorGlombik Do you know an explicit bijection between $\mathbb{R}$ and $\mathscr{P}(\mathbb N)$, e.g.? You can build on that. $\endgroup$ Apr 27, 2019 at 14:31

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