Just a basic pointset topology question: clearly we can detect differences in topologies using convergent sequences, but is there an example of two distinct topologies on the same set which have the same convergent sequences?

$\begingroup$ See also this post at math.SE: Example of different topologies with same convergent sequences $\endgroup$ – Martin Sleziak Dec 4 '16 at 14:52
In a metric (or metrizable) space, the topology is entirely determined by convergence of sequences. This does not hold in an arbitrary topological space, and Mariano has given the canonical counterexample. This is the beginning of more penetrating theories of convergence given by nets and/or filters. For information on this, see e.g.
http://math.uga.edu/~pete/convergence.pdf
In particular, Section 2 is devoted to the topic of sequences in topological spaces and gives some information on when sequences are "topologically sufficient".
In particular a topology is determined by specifying which nets converge to which points. This came up as a previous MO question. It is not covered in the notes above, but is well treated in Kelley's General Topology.

$\begingroup$ I don't recall if Kelley treats this point, but one interesting subject to read a bit on in this context is that of sequential spaces. $\endgroup$ – Mariano SuárezÁlvarez Aug 22 '10 at 16:28

$\begingroup$ @Mariano: I believe Kelley does not, but I do: see Section 2.2. :) $\endgroup$ – Pete L. Clark Aug 22 '10 at 16:43

1$\begingroup$ To be more precise: Kelley's text was written in 1955. The first significant work on sequential spaces (including identifying them by name) was done by S.P. Franklin in 1965. $\endgroup$ – Pete L. Clark Aug 22 '10 at 16:46

$\begingroup$ It MAY be treated in Willard's GENERAL TOPOLOGY,which has a pretty comprehensive treatment of generalized convergence. I'll check and get back to you guys.By the waya basic but good question that a lot of people don't ask when learning the subject,Tony. $\endgroup$ – The Mathemagician Aug 22 '10 at 17:57

1$\begingroup$ As far as I can see, Willard's treatment of sequential convergence is limited to some exercises in Section 10. He does not define or consider sequential spaces per se. $\endgroup$ – Pete L. Clark Aug 22 '10 at 21:12
The cocountable topology on an uncountable set is undistinguishable from the discrete topology if you can only use sequences.
Just another example. Consider the Banach space $\ell^{1}\left(\Gamma\right)$ , $\Gamma$ being an infinite set. Then the weak topology and the norm topology have the same convergent sequences (Schur' Theorem), while they are clearly distinct.

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There is a category of "sequential spaces" in which objects are spaces defined by their convergent sequences and morphisms are those maps which send convergent sequences to convergent sequences.
As stated above, all metric spaces are sequential spaces, but so are all manifolds, all finite topological spaces, and all CWcomplexes.
To build this category, one actually just needs to look at the category of right $M$sets for a certain monoid $M$. Consider first the "convergent sequence space" $S:=${$\frac{1}{n}n\in{\mathbb N}\cup${$\infty$}}$\subset {\mathbb R}$. In other words $S$ is a countable set of points converging to 0, and including $0$. Let $M$ be the monoid of continuous maps $S\to S$ with composition. Then an $M$set is a "set of convergent sequences" closed under taking subsequences.
The category of $M$sets is a topos, so it has limits, colimits, function spaces, etc. And every $M$set has a topological realization which is a sequential space.

$\begingroup$ I'm not sure I'm undertanding you properly. Are the objects in this category merely sequential topological spaces in the sense of Franklin  i.e., topological spaces such that a set $S$ is closed iff every limit of a convergent sequence of elements in $S$ also lies in $S$  or are they something more abstract? $\endgroup$ – Pete L. Clark Aug 22 '10 at 21:51

2$\begingroup$ One reason I ask is that  unless I am mistaken  if $f: X \rightarrow Y$ is a map between topological spaces and $X$ is sequential, then $f$ is continuous iff it preserves convergent sequences, so in particular a candidate definition for the category of sequential spaces would just be the full subcategory of sequential topological spaces and continuous maps. How does this compare to your category? $\endgroup$ – Pete L. Clark Aug 22 '10 at 21:54

1$\begingroup$ I think it's better to look not at the category of all Msets, but the subcategory of sheaves for the canonical Grothendieck topology on M. Those sheaves are called "subsequential spaces" and include the category of sequential topological spaces and continuous maps as a full reflective subcategory. Moreover, any sequentiallyHausdorff subsequential space is a sequential topological space. Cf ncatlab.org/nlab/show/subsequential%20space and P.T. Johnstone's article "On a topological topos". $\endgroup$ – Mike Shulman Aug 23 '10 at 0:13

$\begingroup$ Oh Mike, yes  that's what I meant  it's been a long time since I thought about this. $\endgroup$ – David Spivak Aug 23 '10 at 0:57
I like the following example. Let $D$ be an infinite discrete space and $\beta D$ its StoneČech compactification. Then a sequence in $\beta D$ converges if and only if it is eventually constant. Thus convergent sequences do not distinguish between the compact topology of $\beta D$ and the discrete topology on its underlying set. Obviously  the same holds for every refinement of $\beta D$.
In the space $[0,\omega_1]$ with the order topology the point $\omega_1\in\overline{[0,\omega_1)}$, but for every convergent sequence $(x_n)_{n\in\Bbb N}\subset[0,\omega_1)$, $$\lim_{n\to\infty}x_n<\omega_1.$$