Does every element of the weak-star closure of a set belong to the weak-star closure of a bounded subset? I feel like this must be a monumentally stupid question. Say $X$ is a Banach space, $S\subset X^*$, and $x^*$ is in the weak* closure of $S$. Must $x^*$ lie in the weak* closure of some norm-bounded subset of $S$?
(If $x^*$ is the weak* limit of a sequence of elements of $S$ then this is clear by Banach-Steinhaus. But a convergent net of scalars need not be bounded...)
 A: The Krein-Smulian theorem says that for a convex set $S$, having weak*-closed intersections with closed balls implies being weak*-closed. The fact that convexity is required here suggests that the answer to your question should be negative.  And it is. 
Let $X=\ell^2$. In the dual space, also $\ell^2$, consider the "infinite ellipsoid" 
$$
S = \left\{y : \sum_{n=1}^\infty y_n^2/n^2=1\right\}
$$
Every line through the origin meets $S$; therefore, $0$ is in the weak closure of $S$. 
On the other hand, $0$ is not in the weak closure of any intersection $S\cap B_R$ where $B_R=\{y:\|y\|\le R\}$. Indeed, consider the weak-open sets 
$$
U_N = \left\{y: \sum_{n=1}^N y_n^2 < \frac12\right\}
$$
If $N$ is large enough, then for every $y\in U_N\cap B_R $ we have 
$$
  \sum_{n=1}^\infty y_n^2/n^2 < \frac12+ \frac{R^2}{(N+1)^2} < 1
$$
hence $U_N\cap (B_R\cap S)= \varnothing$. 
A: In addition to the answer by 404, I have two further counterexamples in $\ell^2$:
\begin{align*}
A &= \{ \sqrt{n} \, e_n : n \in \mathbb{N}\} \\
B &= \{ e_m + m \, e_n : m,n \in \mathbb{N}, 1 \le m < n \}
\end{align*}
Again, $0$ is in the weak* closure, but all intersections of $A$ or $B$ with bounded sets are discrete! (It is also interesting to note that this implies that there is no sequence in $A$ or $B$ converging weak*ly to $0$.)
A: The reference to Krein-Smulian in one of the answers is actually a little misleading, and is related to the following warning in Conway's Functional Analysis, page 161:

In fact, you can find a Banach space $X$ and $S\subseteq X^\ast$ a subspace (so certainly convex) such that $S$ is weak$^*$-dense in $X^\ast$, but there is a vector $f_0\in X^\ast$ such that no bounded net in $S$ converges to $f_0$; equivalently, $f_0$ is not in the weak$^\ast$-closure of $S$ intersected with any bounded subset of $X^\ast$.
I do not know a canonical reference for a construction.  However, it is fairly easy to adapt this Mathoverflow answer.  I wrote up the details here (github).
