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In classical topology (dimension theory) one learns that every finite dimensional compact metric space can be embedded into some finite dimensional Euclidean space. Then, by taking the one point compactification of the Euclidean space one gets an embedding of the original compact metric space into a unit sphere. Does the same result hold for infinite dimensional compact metric spaces?

Perhaps a different questions would be the following: We know that we can embed a compact metric space $X$ isometrically into the space of continuous and bounded real valued functions on $X$. Can this (infinite dimensional) Banach space be embedded into the unit sphere of another Banach space?

Any references would be appreciated.

EDIT: By dimension I mean the covering dimension. See the following:

https://en.wikipedia.org/wiki/Lebesgue_covering_dimension

Robert Thingum
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    How do you define 'dimension' in a general metric space? – Berci Sep 25 '19 at 14:54
  • In the case of compact metric spaces most common notions of dimension coincide. One may take the dimension to be the covering dimension. Namely $dim(X)\leq n$ if every finite open cover admits an open finite refinement of order $\leq n+1$. – Robert Thingum Sep 25 '19 at 14:57
  • I just found this question:

    https://math.stackexchange.com/questions/678214/homeomorphisms-between-infinite-dimensional-banach-spaces-and-their-spheres?rq=1

    Which seems to answer my question.

     – Robert Thingum Sep 25 '19 at 14:59
  • If that answers your question, then it would be reasonable to close this question as a duplicate. – Xander Henderson Sep 25 '19 at 15:07
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    The other question does not quite answer yours: it only works for certain classes of spaces. Since you do not insist on the space being embedded into its own sphere, you can give an unconditional positive answer, which is also pretty easy: any real Banach space $X$ embeds into the sphere of $X\times {\mathbf R}$ (e.g. with the Euclidean product norm) via the stereographic projection. – tomasz Sep 25 '19 at 15:29

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