@Perturbative: Balarka was just explaining to you in one word what your mistake was. Lee will talk about group actions on manifolds later. That's different.
@TedShifrin No I didn't mean group actions, I just meant like $v$ acting on $f$, what does that mean? Does it mean something like $\nabla = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle$ "acting on $f$ by $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle$ from typical multivariable calculus
@TedShifrin Well a space is locally compact if every point as a compact neighbourhood (there are plenty of definitions all equivalent to this one in an Hausdorff space so this might not be the best choice). I read that no point in $\ell^2$ (let's start with a nice space) has one such neighbourhood
@TedShifrin well, the simplest point is just that an elliptope is a set of n-by-n correlation matrices (symmetric with diagonal elements equal to one and with nonnegative eigenvanlues)
@PVAL: I'm about to go on Medicare in two months. Who knows what'll happen. I can afford whatever (he says naively), but all the poor old folks who can't ... And all the folks in nursing homes that Medicaid pays for? Trash.
@TedShifrin Oh, of course, we're in a metric space where sequential compactness is compactness! So for example in $\ell^2$ the unit closed ball is not compact because of the sequences with a single $1$ in the $i$-th position
@TedShifrin I ended up asking one of my profs for the convergent thing. Makes sense - rearrange a conditionally convergent series until it does what you want.
I now see why wiki says that one might be tempted to think that all closed compact subsets of $\ell^2$ are finite dimensional and the Hilbert cube is a counterexample
So since no compact set contains a closed ball no compact set contains an open ball either and it's hard for points to have compact neighbourhoods in this situation
The thing which confused me a lot back in the days are manifold with corners. If you're working topologically, they are manifold with boundaries. But if you're working smoothly, they are very not.
I remember how interesting it was to realize that the symmetric products of Riemann surfaces are always (smooth) complex manifolds, but that fails for higher dimension. The symmetric square of $\Bbb C^2$ has interesting singularities.
I actually had to talk about that in some talk I gave as a grad student — I don't remember why.