# Is the dual space of a Banach space Hausdorff?

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

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

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Technically, it is important to say under which topology we are considering the dual space (there are several viable options). I will assume you meant the topology induced by the operator norm. We have the following

Fact: Every metric space $(X, d)$ is a Hausdorff topological space. (Of course we implicitly assume $X$ is given the topology induced by the metric, i.e., the topology consisting of the sets such that each point in the set has a surrounding open ball which is also in the set).

Proof: Suppose $x,y \in X, x \neq y$ and thus $d(x,y) = r > 0$. Then if $z \in B(x,\frac{r}{3})$ we have $$r = d(x,y) < d(x,z) + d(z,y) < r/3 + d(z,y).$$ Thus $d(z,y) > 2r/3$ which shows that $z \notin B(y, \frac{r}{3})$. Essentially the same argument shows that $z \in B(y,\frac{r}{3})$ implies $z \notin B(x, \frac{r}{3})$. Thus $B(x,\frac{r}{3})$ and $B(y,\frac{r}{3})$ are neighborhoods of $x$ and $y$ respectively such that $B(x,\frac{r}{3}) \cap B(y,\frac{r}{3}) = \emptyset$. Since $x,y$ were arbitrary this shows that $(X,d)$ is Hausdorff.

Now notice that the dual of a Banach space is a normed space (in fact also a Banach space, but this is irrelevant) because it is normed by the operator norm $$\|x^*\|_{X^*} = \sup_{x \in X}|x^*(x)|.$$

This norm induces a metric in the usual way, and thus $X^*$ is 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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