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Show that any compact metric space $X$ can be isometrically embedded into $C([0,1])$, the space of continuous functions over $[0,1]$ with sup-norm $(||f||_\infty = sup_x(|f(x)|)$

I have no idea yet how to approach this, any help would be appreciated.

cmk
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Ilan Aizelman WS
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    What kind of results have you learned so far that might help you out here? It's important to share your own thoughts when asking questions on MSE. – Theo Bendit Jul 12 '19 at 12:25
  • @TheoBendit Well, that space $C[0,1]$ with the sup-norm should be Banach. Also, I believe the intuition is to show that any sequence in $X$ has a convergent subsequence, and thus it'll be compact. – Ilan Aizelman WS Jul 12 '19 at 12:43
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    Idea: start by considering a countable dense subset of $X$. If you are able to embed it, then all other elements of $X$ will be defined by continuity. To define the embedding on a countable set, you can proceed by induction, using the completness of $C([0,1])$ and the fact that it's infinite dimensional. – Crostul Jul 12 '19 at 13:24
  • Bessaga and Pelczynski (topics in infinite-dimensional topology), chapter II, paragraph 1, p. 49-51 has a proof. – Henno Brandsma Jul 12 '19 at 21:57
  • The proof in B-P holds for all separable metric spaces. Compactness is a red herring here. The image is linearly independent (and closed iff $X$ is complete, of course). – Henno Brandsma Jul 12 '19 at 22:24
  • @HennoBrandsma this book is difficult to find... – Gabriel Romon Jul 13 '19 at 07:39
  • I just walk to my cupboard and take it out. It's a classic... – Henno Brandsma Jul 13 '19 at 07:42
  • This is the Banach--Mazur theorem. https://hsm.stackexchange.com/questions/3728/where-was-the-banach-mazur-theorem-originally-published https://en.wikipedia.org/wiki/Banach%E2%80%93Mazur_theorem – Tomasz Kania Jul 13 '19 at 09:19
  • @TomekKania Banach-Mazur is about embedding Banach spaces, not general metric spaces. One more step is needed. – Henno Brandsma Jul 14 '19 at 06:47

1 Answers1

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If $(X,d)$ is compact metric, $C(X)$ in the sup norm is a Banach space.

Fix $p \in X$. For $x \in X$ define $f_x: X \to \Bbb R$ by $f_x(y)=d(y,x)-d(y,p)$. This $f_x$ is well-defined and continuous, as the metric is a continuous function.

Now check that $F: X \to C(X)$ defined by $F(x)=f_x$ is an isometric embedding.

So $X$ embeds isometrically into $C(X)$, and a classical fact is that $X$ (being compact metric) is a continuous image of the Cantor set $2^\omega$, so we have $\phi: 2^\omega \to X$ a continuous surjection and this induces an isometric injection $\phi^\ast: C(X) \to C(2^\omega): \phi^\ast(f)=f \circ \phi$, another classic fact.

Moreover $C(2^\omega)$ embeds isometrically into $C([0,1])$ by linear interpolation, essentially. This is also well-known.

So we can combine $$X \hookrightarrow C(X) \hookrightarrow C(2^\omega) \hookrightarrow C([0,1])$$ as a chain of isometric embeddings.

Henno Brandsma
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