property of a topological group

This should be easy, but apparantly not for me. Let G be a topological group, and let $\mathcal{N}$ be a neighbourhood base for the identity element $e$ of $G$. Then for all $N_1,N_2 \in \mathcal{N}$, there exists an $N' \in \mathcal{N}$ such that $e \in N'\subset N_1 \cap N_2.$

This means for example that every $N \in \mathcal{N}$ contains the identity element, which seems strange to me.

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It's not a specific property of topological groups, since it's true by definition of a neighborhood basis of any element in a topological space. (but you can construct a basis of symmetric neighborhood of the identity element, if you want to use the fact that you work in a topological group) – Davide Giraudo Oct 29 '11 at 21:29
That's not strange at all -- if you have some point, and a neighborhood basis for it, every element of this neighborhood basis must contain the point. That's nothing to do with groups. (Indeed, nothing in this problem has anything to do with groups; this holds for G an arbitrary topological space and e an arbitrary point.) – Harry Altman Oct 29 '11 at 21:30
@Nadori: There's no need to delete, someone might have the same confusion sometime and we'd want this to be up to help them out. – Zev Chonoles Oct 30 '11 at 21:52

Given a point x in a topological space X and a neighborhood basis $\mathcal{N}$ for x, and $N_1, N_2 \in \mathcal{N}$, since $N_1, N_2$ are in this neighborhood basis, they are, in particular, neighborhoods of x. (And, note, therefore contain x.) Thus so is their interesection. And any neighborhood of x must contain some element of $\mathcal{N}$ (which in turn contains x, as they all do), because that's what it means for $\mathcal{N}$ to be a neighborhood basis.
Milne uses another definition: Recall that a neighbourhood base for a point x of a topological space X is a set of neighbourhoods $\mathcal{N}$ such that every open subset U of X containing x contains an N from $\mathcal{N}$. So the above definition does not require the containment of x by every element of $\mathcal{N}$. See also planetmath.org/encyclopedia/BaseOfANeighborhoodSystem.html Your definition makes everything trivial, of course. – Nadori Oct 29 '11 at 22:41
Wait, by "neighborhoods" do you just mean "sets containing a nonempty open set" rather than "neigbhorhoods of x"? If so, this is false; by that definition, $(-a,0)\cup(0,a)$ could be included in a neighborhood basis of 0 in R, despite not containing 0. I think you really mean to say "neighborhoods of x" (making it trivial, yes). – Harry Altman Oct 29 '11 at 23:54
Neighborhoods are by definition open sets. Well I'll just interpret it as a set of neighborhoods of $x$. Confusion is gone :) – Nadori Oct 30 '11 at 10:45