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I Separable Vector space

  1. Jan 22, 2017 #1
    Dear forum,

    I am trying to understand what a separable vector space is. I know we can perform the tensor product of two or more vector space and obtain a new vector space. Is that vector space separable because it is the product of other vector spaces?

    thanks
     
  2. jcsd
  3. Jan 23, 2017 #2

    Orodruin

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    No. A separable space contains a countable dense subset. The product of two vector spaces will not be separable unless the two original vector spaces were.
     
  4. Jan 23, 2017 #3

    micromass

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    The notion of a vector space alone is not sufficient for separability to be well defined.
     
  5. Jan 23, 2017 #4
    Thank you.

    What would a countable dense subset be? I understand it contains a subset of vectors that has a particular property. Could you give me an example of what a countable dense subset means from ordinary finite dimensional linear algebra? Let's consider the vector space of vectors
    ##v=(v_{x} , v_{y}, v_{z})##...
     
  6. Jan 23, 2017 #5

    micromass

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    Dense makes no sense on a vector space only. You'll need further structure, like a norm, a distance or a topology.

    For example, the vector space ##\mathbb{R}^3## can be equipped with the norm
    [tex]\|v\| = \sqrt{v_x^2 + v_y^2 + v_z^2}[/tex]
    A countable dense set can then be given by [tex]\{v~\vert~v_x, v_y,v_z\in \mathbb{Q}\}[/tex]

    But let's start with the basics, do you know what a norm is?

    Also, where did you encounter the notion of separable? Seeing the context might help.
     
  7. Jan 23, 2017 #6

    Orodruin

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    Obviously. This is implied by the "dense". Without some sort of topology, this is not well defined.
     
  8. Jan 23, 2017 #7

    micromass

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    You know that. I know that. The point is that the OP might not.
     
  9. Jan 23, 2017 #8
    Hi micromass,

    The norm is, conceptually, the "length" of a vector. I run into the idea of separable vector space in introductory quantum mechanics where Hilbert vector space is said to be separable. this leads down to the discussion of such a space being a tensor product of other vector spaces...
     
  10. Jan 23, 2017 #9

    micromass

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    OK, since it's in the context of Hilbert spaces, I can give a simpler definition of separable. All it means is that there is a countable orthonormal basis. That is: there is a subset of the Hilbert space ##\{e_n~\vert~n\in I\}## with ##I## countable, which intuitively means ##I## finite or ##I=\mathbb{N}## such that
    1) ##\|e_n\| = 1## for all ##n##
    2) ##<e_n, e_m> = 0## for ##n\neq m##
    3) Every ##x## in the Hilbert space can be written as ##x = \sum_{n\in I}\alpha_n e_n## for some ##\alpha_n\in \mathbb{C}## (it can be shown that ##\alpha_n =<x,e_n>##). Note this sum is an infinite sum (=series) when ##I=\mathbb{N}## and convergence comes into play, which makes it distinct from linear algebra where all sums are finite.
     
  11. Jan 23, 2017 #10

    micromass

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    As an example in ##\mathbb{C}^3##, take ##(1,0,0)##, ##(0,1,0)##, ##(0,0,1)##. These are orthonormal and form a basis.
     
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