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Is there any counterexamples for that? or there is quick proof?

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In what topology? –  Qiaochu Yuan Oct 22 '12 at 13:53

3 Answers 3

The dual of Banach space is a Banach space, hence Hausdorff.

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Here:

normed => metric => Hausdorff

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For $X^*$ to be Hausdorff in the weak topology, or even the weak* toplogy, it is enougn to know that the elements of $X$ separate points of $X^*$. But that is just the definition of equality for functions.

Given two distinct points $f,g \in X^*$, there is $x \in X$ so that $f(x) \ne g(x)$, and from this we get two disjoint open neighborhoods $\{k \in X^*: k(x) > r\}$ and $\{k \in X^*: k(x) < r\}$ where $r$ is chosen between $f(x)$ and $g(x)$.

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