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.

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    $\begingroup$ 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. $\endgroup$ – Theo Bendit Jul 12 '19 at 12:25
  • $\begingroup$ @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. $\endgroup$ – Ilan Aizelman WS Jul 12 '19 at 12:43
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    $\begingroup$ 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. $\endgroup$ – Crostul Jul 12 '19 at 13:24
  • $\begingroup$ Bessaga and Pelczynski (topics in infinite-dimensional topology), chapter II, paragraph 1, p. 49-51 has a proof. $\endgroup$ – Henno Brandsma Jul 12 '19 at 21:57
  • $\begingroup$ 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). $\endgroup$ – Henno Brandsma Jul 12 '19 at 22:24

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.

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  • $\begingroup$ Nice. Can simply use $f_x(y)=d(x,y)$ as well, right? $\endgroup$ – Tommy1234 Jul 29 '19 at 7:50
  • $\begingroup$ @Tommy1234 yes for compact $X$ l think. $\endgroup$ – Henno Brandsma Jul 29 '19 at 7:53
  • $\begingroup$ Thank you lot lot lot. $\endgroup$ – Sebastiano Aug 21 at 12:19

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