Just a basic point-set 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?
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$\begingroup$ See also this post at math.SE: Example of different topologies with same convergent sequences $\endgroup$– Martin SleziakCommented Dec 4, 2016 at 14:52
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6 Answers
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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://alpha.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.
answered Aug 22, 2010 at 16:19
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$\begingroup$ 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. $\endgroup$ Commented Aug 22, 2010 at 16:28
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$\begingroup$ @Mariano: I believe Kelley does not, but I do: see Section 2.2. :) $\endgroup$ Commented Aug 22, 2010 at 16:43
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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. $\endgroup$ Commented Aug 22, 2010 at 16:46
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$\begingroup$ 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 way-a basic but good question that a lot of people don't ask when learning the subject,Tony. $\endgroup$ Commented Aug 22, 2010 at 17:57
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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. $\endgroup$ Commented Aug 22, 2010 at 21:12
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The cocountable topology on an uncountable set is undistinguishable from the discrete topology if you can only use sequences.
answered Aug 22, 2010 at 16:06
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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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18$\begingroup$ +1 for an example "in nature". $\endgroup$ Commented Aug 22, 2010 at 20:38
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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 CW-complexes.
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.
answered Aug 22, 2010 at 21:09
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$\begingroup$ 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? $\endgroup$ Commented Aug 22, 2010 at 21:51
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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? $\endgroup$ Commented Aug 22, 2010 at 21:54
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1$\begingroup$ I think it's better to look not at the category of all M-sets, 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 sequentially-Hausdorff subsequential space is a sequential topological space. Cf ncatlab.org/nlab/show/subsequential%20space and P.T. Johnstone's article "On a topological topos". $\endgroup$ Commented Aug 23, 2010 at 0:13
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$\begingroup$ Oh Mike, yes -- that's what I meant -- it's been a long time since I thought about this. $\endgroup$ Commented Aug 23, 2010 at 0:57
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In the space $[0,\omega_1]$ with the order topology the point $\omega_1\in\overline{[0,\omega_1)}$, but for every convergent sequence $(x_n)_{n\in\Bbb N}\subset[0,\omega_1)$, $$\lim_{n\to\infty}x_n<\omega_1.$$
answered Dec 10, 2014 at 8:42
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I like the following example. Let $D$ be an infinite discrete space and $\beta D$ its Stone-Čech 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 $\beta D$.
answered Dec 10, 2014 at 9:25