One-Dimensional Subsets of Hilbert Space vs. Euclidean Space

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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 $mathbbR^3$?



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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    up vote
    1
    down vote

    favorite












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



    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!







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











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



      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!







      share|cite|improve this question












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



      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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      share|cite|improve this question




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      asked Aug 22 at 7:40









      John Samples

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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 $mathbbR^2n+1$.






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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            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 $mathbbR^2n+1$.






            share|cite|improve this answer
























              up vote
              1
              down vote



              accepted










              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 $mathbbR^2n+1$.






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                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 $mathbbR^2n+1$.






                share|cite|improve this answer












                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 $mathbbR^2n+1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 22 at 9:11









                John Samples

                1,031516




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