Did I mention my finals are now over? I will soon be more drunk than you can possibly imagine. Here's a question people often ask me: if X is an infinite-dimensional Banach space, why is any nonempty open set in the weak^* topology on X^* unbounded in the norm topology?

Good question! Non-math-dorks may wish to look elsewhere for a moment. It gets hairy.

The weak^* topology on X^* is generated by sets of the form U_xyr={f: |f(x)-y|‹r} for x in X, y in our field K (either the reals or complex numbers), and r›0. Thus any open set is a union of finite intersections of such sets. So if any finite intersection is unbounded, any open set will be unbounded. What do the finite intersections look like? They are sets of the form O={f: |f(x_i)-y_i|‹r_i, i=1...n}. That is, we pick out a finite number of vectors, and our set consists of all linear functionals whose value at those vectors is 'sufficiently close' to the value we specify. Since there are only a finite number of these vectors, and X is itself infinite-dimensional, we can choose a vector x in X linearly independent from x₁...x_n; normalise it, so that x has norm 1. The linear subspace N spanned by the x_i is finite-dimensional, hence closed (by a pretty argument I may or may not share with you later on). Then x does not lie in N. Hence by a corollary of the Hahn-Banach Theorem, we can find a functional f in X^* with f(y)=0 for all y in N, but f(x)=1. Now, if there is some f₀ in O, let us construct a new functional g in X^* by letting, for any M›0, g=f₀+(M-f₀(x))f. Then since g agrees with f₀ on N, g certainly lies in O, yet g(x)=f₀(x)+Mf(x)-f₀(x)f(x)=Mf(x)=M. Thus ||g|| is at least M, since ||g||=sup{|g(z)|: ||z||=1}. Since M was arbitrary, O must be unbounded. QED.