# Show that any compact metric space $X$ can be isometrically embedded into $C([0,1])$, the space of continuous functions over $[0,1]$

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.

• 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
• 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

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.

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