Why is the sequential criterion true in metric spaces? When does it fail in general? I’ve read just the basics of some introductory analysis books and sometimes they show that we can characterize things like limits, continuity, compactness, etc. in terms of sequences. 
I’ve heard that these sequential criteria hold for general metric spaces, but that in topology for example one encounters situations where sequences aren’t quite sufficient, or where it’s better to consider some other object.
My questions are: 


*

*Is there some intuition for why the sequential criterion holds in things like Euclidean space or general metric spaces, but not in some other spaces? 

*Does it simply have to do with the fact that we have a metric, and if so, why does the metric “induce” such sequential criteria (versus without a metric we may not)? 

*Is the notion of distance/metric captured in some way by sequences because approaching some value is equivalent to a sequence approaching that value? 

*Are there any ways by which we can determine whether a general space possesses these sequential criteria? They seem quite useful.

 A: Many of these hold for sequential spaces. These can be defined in a variety of equivalent ways. One simple way that uses no new terminology is that $X$ is sequential iff for each non-closed $A\subseteq X$ there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ converging to a point of $(\operatorname{cl}A)\setminus A$. It turns out that this is also equivalent to the statement that continuity of functions on $X$ is determined by sequences: $X$ is sequential iff for every space $Y$ a function $f:X\to Y$ is continuous iff it preserves convergent sequences, i.e., iff $\langle f(x_n):n\in\Bbb N\rangle$ converges to $f(x)$ in $Y$ whenever $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in $X$.
Sequential compactness and countable compactness are equivalent in sequential space, but unlike the situation in metric spaces, they are not equivalent to compactness: the space of countable ordinals with the linear order topology is first countable, hence sequential, and both countably and sequentially compact, but it is not compact.
If $X$ is second countable (i.e., has a countable base for the topology), then it is compact iff it is sequentially compact, as shown in the answer to this question, but that is more than is needed; for instance, the comments under the question show that they two are equivalent in Lindelöf Hausdorff spaces. (Every second countable space is sequential and Lindelöf, but a sequential Lindelöf space need not be second countable.) The comments also note, with a reference, that these types of compactness are equivalent for the weak topology on Banach spaces, which is a sequential only if the space is finite-dimensional.
A: James R. Munkres, in Topology: A First Course  section 3-7, defines two reasonable weakenings of compactness. On page 178, limit point compactness is when every infinite subset of the space has a limit point. On page 179, when every sequence has a convergent subsequence, he calls it sequential compactness. It turns out that the product of two sequentially compact spaces is again sequentially compact. However, the product of two limit point compact spaces need not be limit point compact. An example is item 112 in Counterexamples in Topology by Steen and Seebach.
