Exact meaning of a local base at zero in a topological vector space...

  1. I am confused as to exactly what a local base at zero (l.b.z.) tells us about a topology. The definition given in Rudin is the following: "An l.b.z. is a collection G of open sets containing zero such that if O is any open set containing zero, there is an element of G contained in O". Ok, great.

    But I have seen some proofs in my functional analysis class that suggest something like the following: Any open set in the topology can be formed by taking unions (possibly uncountable) of *translations* of sets in a l.b.z. Is this true, or am I just missing something?
  2. jcsd
  3. micromass

    micromass 18,426
    Staff Emeritus
    Science Advisor

    Yes, the two are equivalent!

    Basically, take an open set G in the topology. If a is in G, then G-a contains 0, thus we can find an element V_A of the lbz, such that [tex]V\subseteq G-a[/tex]. Thus [tex]a+V[/tex] contains a and is smaller than G. Now, we can write G as

    [tex]G=\bigcup_{a\in G}{a+V_a}[/tex]

    Thus we have written G as union of translations of the lbz...
  4. Great, thanks! Now that I know that, I'm going to try to work out a proof. But is this discussed in Rudin, or on the web, somewhere in case I get stuck?
  5. micromass

    micromass 18,426
    Staff Emeritus
    Science Advisor

    Sorry I posted too fast. I was going to include a proof. I've edited my post 1 with the proof...
  6. Landau

    Landau 905
    Science Advisor

    You are already familiar with a neighbourhood base in any topogical space.

    Now, the topology on a t.v.s. (or a topological group for that matter) is translation-invariant. This is because "translation by a fixed g"
    [itex]T_g:x\mapsto x+g[/itex]
    is a homeomorphism (which is because addition is by definition continuous, and T_g is obviously invertible). So it suffices to consider the neighborhood base of any point, in particular 0.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook