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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 $T_{1}=T_{2}$?

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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: math.stackexchange.com/questions/53236/… –  Nate Eldredge Jun 12 '12 at 22:33

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up vote 4 down vote accepted

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.

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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 $D\subset X$? –  Thomas E. 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

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