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Exact meaning of a local base at zero in a topological vector space... 
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#1
Mar1811, 02:04 PM

P: 529

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
Mar1811, 02:08 PM

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P: 18,330

Yes, the two are equivalent!
Basically, take an open set G in the topology. If a is in G, then Ga contains 0, thus we can find an element V_A of the lbz, such that [tex]V\subseteq Ga[/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... 


#3
Mar1811, 02:11 PM

P: 529




#4
Mar1811, 02:14 PM

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P: 18,330

Exact meaning of a local base at zero in a topological vector space...
Sorry I posted too fast. I was going to include a proof. I've edited my post 1 with the proof...



#5
Mar1811, 03:44 PM

Sci Advisor
P: 905

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 translationinvariant. 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. 


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