# Collection of converging sequences determines the topology?

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

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.
What role does $X$ play in the above construction. Is $D\subset X$? – T. Eskin Jun 12 '12 at 22:07
@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