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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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    Closely related question with examples that can be adapted: http://math.stackexchange.com/questions/53236/continuity-and-image-of-convergent-sequences?rq=12012-06-12

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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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    @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)2012-06-13