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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)2011-10-29
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    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.)2011-10-29
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    @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.2011-10-30

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