Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
In what topology? – Qiaochu Yuan Oct 22 '12 at 13:53

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

share|cite|improve this answer


normed => metric => Hausdorff

share|cite|improve this answer

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 $g(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.