I'm working on this question:
Let $S$ be a subset of a normed linear space $X$. Suppose for each $l\in X^*$, there exists $M_l$ such that $l_x \leq M_l$ for all $x\in S$. Show that $S$ is bounded in $X$.
The standard solution would be to treat the dual space as the domain and apply the uniform boundedness principle, then making use of the fact that $X$ looks like $X^{**}$ to obtain our conclusion. This suggests that completeness of the base field plays an important role here.
However, I tried to prove the question more directly. Assume that $S$ is unbounded. Pick a sequence $\{s_n\}_{n\in\mathbb{N}}$ such that $||s_n||>n$. Consider $W:= span(s_n|n\in\mathbb{N})$. If $W$ is finite dimensional I managed to show that we can find $l\in X^*$ so that $l(x)$ is unbounded. The problem comes when $W$ is infinite-dimensional.
Suppose $W$ is infinite-dimensional. Then I can extract a linearly independent subsequence of vectors $\{s_{n_k}\}$. Define a linear functional on this subsequence by $l(s_{n_k}) = n_k$. If I can show that $l$ is a bounded linear functional, we are done. Here it starts to get suspicious. As $||s_{n_k}||>n_k$ by our choice, $l(\frac{s_{n_k}}{||s_{n_k}||}<1)$, at the very least. However, I understand that this doesn't mean that every unit vector will be bounded, because the number of components in the expression of it can get arbitrarily large. It seems plausible though, that given a fixed norm, we can redefine $l(s_{n_k}) = f(n_k)$ for some $f$ which diverges sufficiently slowly so that $l$ is a bounded linear functional, and $l(s_{n_k})$ still increases without bound as $n_k\rightarrow\infty$.
Take for instance, when $X$ is the infinite-dimensional Euclidean space equipped with the 1-norm. Here we have no problems letting $l(s_{n_k}) = n_k$, for any unit vector $v$ is writeable as a sum of its components with coefficients summing up to 1, so that $l(v)<1$. Hence $l$ is bounded. It doesn't occur to me why we can't adjust $l$ accordingly as we move to arbitrary norms. Is this where the completeness property subtlely sneaks in?
Much appreciated!