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Finite Dimensional Hausdorff Topological Space

  1. Apr 14, 2012 #1
    How do I prove that a Hausdorff topological space E is finite dimensional iff it admits a precompact neighborhood of zero?
  2. jcsd
  3. Apr 15, 2012 #2


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    i think you mean banach space, not hausdorff space, and then this is a well known theorem of riesz: th.5.9.4., p.109, of dieudonne's foundations of modern analysis.
  4. Apr 15, 2012 #3
    Hi mathwonk,

    I mean Hausdorff space.
  5. Apr 15, 2012 #4
    It is indeed true that a locally compact Hausdorff topological vector space E is finite dimensional.

    Proof: Let K be a compact neighborhood of 0. We can assume that K is balanced. Since (1/2)K is a neighborhood of 0, there are finitely many points x1,...,xn such that

    [tex]K\subseteq (x_1+\frac{1}{2}K)\cup...\cup (x_n+\frac{1}{2}K)[/tex]

    Let M be the finite dimensional subspace spanned by the x1,...,xn. Then M is closed. The quotient space E/M is Hausdorff. Since [itex]K\subseteq M+\frac{1}{2}K[/itex], then [itex]\varphi(K)\subseteq \frac{1}{2}\varphi(K)[/itex]. So (by induction) [itex]\varphi(2^nK)\subseteq \varphi(K)[/itex].

    K is balanced, so [itex]E=\bigcup_n 2^n K[/itex]. Thus [itex]\varphi(E)=\varphi(K)[/itex]. Thus E/M is compact. Which implies that E/M is one point. Thus E=M.
  6. Apr 15, 2012 #5


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    what do you mean by zero in a topological space? oh i see micromass, you assumed he meant a vector space. the argument you gave is a variation of the argument for Riesz's theorem that I referred to. That's a nice example of extending an argument to a more general setting. I didn't know that version.
    Last edited: Apr 15, 2012
  7. Apr 17, 2012 #6
    Thank you micromass. Your proof makes sense.
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