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

by AxiomOfChoice
Tags: base, exact, local, meaning, space, topological, vector
 P: 507 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?
 PF Patron Sci Advisor Thanks Emeritus P: 15,671 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 $$V\subseteq G-a$$. Thus $$a+V$$ contains a and is smaller than G. Now, we can write G as $$G=\bigcup_{a\in G}{a+V_a}$$ Thus we have written G as union of translations of the lbz...
P: 507
 Quote by micromass Yes, the two are equivalent!
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?

PF Patron
 Sci Advisor P: 904 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" $T_g:x\mapsto x+g$ 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.