Collection of converging sequences determines the topology?

Question

Is it the case that the set of converging sequences uniquely determines the open sets in a topological space?

In other words: Given a space $X$ and two topologies $T_{1}$, $T_{2}$ on $X$. such that the set of converging sequences under $T_{1}$ equals the set of converging sequences under $T_{2}$. Does it imply that $T_{1}=T_{2}$?

This is not true in general, but is true for a class of spaces called sequential spaces. — Alex Becker, Jun 12 '12 at 21:52
Closely related question with examples that can be adapted: http://math.stackexchange.com/questions/53236/continuity-and-image-of-convergent-sequences?rq=1 — Nate Eldredge, Jun 12 '12 at 22:33

Brian M. Scott · Accepted Answer · 2012-06-12T22:16:08.823

4

No. A simple counterexample can be produced as follows. Let $D$ be an uncountable set, and fix a point $p\in D$. Let $\tau_1$ be the discrete topology on $D$. Let $\tau_2$ be the topology that makes each point of $D\setminus\{p\}$ isolated and gives $p$ nbhds of the form $D\setminus C$, where $C$ is any countable subset of $D\setminus\{p\}$. In other words, $$\tau_2=\Big\{\{x\}:x\in D\setminus\{p\}\Big\}\cup\{D\setminus C:C\subseteq D\setminus\{p\}\text{ and }C\text{ is countable}\}\;.$$

Then $\langle D,\tau_1\rangle$ and $\langle D,\tau_2\rangle$ have the same convergent sequences: the only sequences that converge in either topology are those that are eventually constant. However, the two topologies are clearly not homeomorphic.

Spaces whose topologies are completely determined by their convergent sequences are called sequential spaces.

edited Jun 12 '12 at 22:16

answered Jun 12 '12 at 21:56

Brian M. Scott

616,228

1

You're getting a lot of mileage from this space today ^^ – Olivier Bégassat Jun 12 '12 at 21:59
@Olivier: Yes, that thought had occurred to me, too. :-) – Brian M. Scott Jun 12 '12 at 22:03
What role does $X$ play in the above construction. Is $D\subset X$? – T. Eskin Jun 12 '12 at 22:07
For my own understanding, the result is true if we replace sequences with nets, correct? The issue with sequences essentially being one of the topological space's first countability (or rather, lack thereof)? – J. Loreaux Jun 12 '12 at 22:08
@Thomas: There is no $X$ in my construction: the space is $D$. If you want to call it $X$ instead, just rename it. – Brian M. Scott Jun 12 '12 at 22:10
@J.Loreaux: Yes, it’s true with nets (or filters). First countability is actually more than you need: sequential spaces are precisely the quotients of first countable spaces, and they aren’t all first countable. – Brian M. Scott Jun 12 '12 at 22:11
@Brian: Thanks, that makes sense. – J. Loreaux Jun 12 '12 at 22:12
@BrianM.Scott: You define $\tau_{2}$ by using the notion of $X$ and $D$ simultaneously. That is the reason why I'm asking. – T. Eskin Jun 12 '12 at 22:15
@Thomas: Typo. I didn’t even notice it when you asked the first time. Fixed now; thanks. – Brian M. Scott Jun 12 '12 at 22:16
No problem. There is still one $X$ left though :-) – T. Eskin Jun 12 '12 at 22:18
@Thomas: Not any more! :-) – Brian M. Scott Jun 12 '12 at 22:19
How about the set of all continuos functions from X to X? Do they determine the topology uniquely? – Shay Moran Jun 12 '12 at 22:24
@Brian: I dont see why it is the case. Consider X=\mathbb{N}. and let \mathcal{P}={{0,2,4,...},{1,3,5,7,.....}} – Shay Moran Jun 13 '12 at 09:33
Shit :)
Consider the mapping 2->0, 4->2, 6->4,..... 1->1, 3->3,5->5,...0->1 The preimage of the even numbers (which is an open set) is the even numbers without 0 (which is not open)
– Shay Moran Jun 13 '12 at 09:35
@Shay: True enough; I’m not sure now what I was thinking. (Possibly I wasn’t: I’d just come back from a hilly 36-mile bicycle ride.) I’ll try to find some time to think about it again. – Brian M. Scott Jun 13 '12 at 09:44
@BrianM.Scott: I hope you enjoyed the ride :)
I will post this question - perhaps more people will see it (My feeling is that the answer is yes)
– Shay Moran Jun 13 '12 at 10:37

Collection of converging sequences determines the topology?

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