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Motivated by a question I received (though a bit different), is there a one-dimensional subset of Hilbert Space that cannot be embedded in $\mathbb{R}^3$?

Here, dimension refers to covering or inductive dimension. I know by the Menger-Nobeling Theorem that any compact, one-dimensional subset of Hilbert Space can be embedded in three-space, but compactifications of Hilbert Space are not something I've studied.

It feels like it might be equivalent to ask if the Hilbert Space and Hilbert Cube can be distinguished by their one-dimensional subsets, as well. Curious to know the answer to either! Thanks!

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The answer is no. It is a generalization of the above-stated Menger-Nobeling Theorem, given as theorem V.3 in Hurewicz/Wallman, that the assumption of compactness may be dropped. In other words, any separable metric space of dimension $n$ can be embedded in $\mathbb{R}^{2n+1}$.

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