Compact subsets of function spaces, geometry The subset is called compact when every open cover contains a finite subcover. In Euclidean spaces, it is easy to visualize this by imagining some open ball that contains this set, thinking about the subset of this ball that contains the set and divising this subset into the finite collection of sets. Is there a way of some kind of visualization in the case of infinite-dimensional function spaces?
 A: There is a generalization of metric space, called continuity space, where the codomain of the metric is replaced by a quantale (a certain partially ordered set with some extra structure, see Flagg's "quantales and continuity spaces"). It it then shown that every topological space is metrizable as long as one interprets metrizable to mean with respect to a continuity space structure. That means that all of the geometric intuition valid for metric spaces can be carried over to arbitrary topological spaces. 
A: The topologies for infinite dimensional Banach spaces can be a bit tricky. One nice "measure" of their weird behavior with respect to compactness is looking at the unit ball (vectors of norm $\leq 1$). The unit ball in a Banach space is compact (with respect to the norm topology) if and only if the Banach space is finite dimensional. In some sense, there are too many open sets. One of the classical results of functional analysis is the Banach-Alaoglu theorem, which states that if $X^*$ is the dual space of some Banach space, then its unit ball is compact in the weak-$*$ topology. (As a side note, in this kind of weak topology for an infinite dimensional Banach space, the (nonempty) open sets are all unbounded. ) In my experience working with weak topologies, I think the best way of thinking about compactness is in terms of nets. More precisely, a (topological) space is compact iff every net has a convergent subnet. This is because weak convergence is easier to check than finite subcoverings. Suppose $X$ is Banach and $Y$ a family of functionals on $X$, and $X$ is given the $Y$-weak topology. Then $x_{\alpha} \rightarrow x$ weakly if $l(x_{\alpha}) \rightarrow l(x)$ in $\mathbb{C}$ for all $l\in Y$.  
The source I used for functional analysis is Reed, Simon's book Functional Analysis Vol I. 
A: If your metric space is complete you have that a set is totally bounded if and only if it is relatively compact.
A set a totally bounded if for every $\epsilon > 0$ there is a finite $\epsilon$-net. 
A set is relatively compact if the closure is compact. 
This somehow can bring your intuition from the finite dimensional case to the infinite dimensional case, when your space is complete. 
