Convergence with respect to two topologies I have an exercise in general topology as follows: Let $T$ and $T'$ be two topologies in a set $X$. What condition we need to put on these two topologies so that, if $(x_n)$ converges to $x$ with respect to the topology $T$, then $(x_n)$ also converges to $x$ with respect to the topology $T'$?
I guess the condition should be: for every $U'\subset T'$ and every $x\in U'$, there exists $U\subset T$ such that $x\in U\subset U'$. Of course, this condition is sufficient, but I do not know if it is also necessary.
Someone can help me? Thanks for any help!
 A: It is not necessary.  On an uncountable set $X$, let $T'$ be the discrete topology and $T$ the cocountable topology.  Note that all convergent sequences in either topology are eventually constant.
EDIT: A necessary and sufficient condition is (A) and (B):
(A).   For any $x, y$ and $U \in T'$ containing $x$ but not $y$, there is $V \in T$ containing $x$ but not $y$.
(B).  For any countably infinite set $S$, any $U \in T'$ disjoint from $S$ and $x \in U$, there exists $V \in T$ with $x \in V$ and
$S \backslash V$ infinite. 
Of course (A) is automatically true if $(X,T)$ is a $T_1$ space.
Necessity:  Suppose (A) fails.  Taking the sequence $x_n = y$, we have $x_n \to x$ in $T$ but not in $T'$.
On the other hand, suppose (B) fails, so there exist $S, U, x$ as above with no such $V$.  Let $x_n$ be an enumeration of $S$.  Then $x_n \to x$ in $T$ but not in $T'$.
Sufficiency: Suppose $x_n \to x$ in $T$ but not in $T'$.
Thus there is $U \in T'$ with $x \in U$ but infinitely many $x_n \notin U$.
Taking a subsequence, we may assume all $x_n \notin U$.
Now if there is some $y$ that occurs infinitely often in the sequence, 
$y \notin U$, so by (A) there is $V \in T$ containing $x$ but not $y$, 
contradicting  $x_n \to x$ in $T$.
On the other hand, if there is no $y$ that occurs infinitely often, 
the set $S = \{x_n: n \in \mathbb N\}$ is countably infinite and disjoint from $U$, so by (B) there is $V \in T$ with $x \in V$ and infinitely many $x_n \notin V$, again contradicting $x_n \to x$ in $T$.
