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Can someone give me a proof or a counterexample to the following: Given two different topologies over the same set, the neighborhood bases of these two topologies have to differ in at least one point.

(I tried doing a proof by contradiction but couldn't find it)

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Try contraposition: If the neighborhood bases are the same everywhere, then the topologies are identical. – Henning Makholm Sep 9 '11 at 17:05
According to Petersen's "Analysis now", Exercise 1.2.1, Hausdorff defined topology in terms of assigning to $x\in X$ a nonempty family of subsets that satisfied the relevant conditions (each contains $A$; the intersection of any two contains a third element of the family; and for every $A$ in the family for $x$ and every $y\in A$, there is a $B$ in the family corresponding to $y$ that is contained in $A$). The exercise in question is to show that such a family of families uniquely determines the topology. – Arturo Magidin Sep 9 '11 at 18:14

You can recover the topology from knowledge of a neighbourhood basis: a set is open if and only if it contains a basic neighbourhood of each of its points. Thus two topologies having identical neighbourhood bases are identical.

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