Suppose $(a_n)$ is a sequence in $X$ on which there are defined two topologies $T_1,T_2$. If $(a_n)$ converges to a unique limit $x_1$ wrt $T_1$, and to a unique limit $x_2$ wrt $T_2$, is it necessary that $x_1=x_2$? If not are there any nice counter-examples?
I have tried to show that this is false using a counter-example, by "interchanging" two points in $X$. Take $X=\mathbb{R}$ and $T_1$ to be the standard topology on $\mathbb{R}$, so that the sequence $(2^{-n})$ converges uniquely to $0$ wrt $T_1$. Now "interchange" $1$ and $0$ when defining $T_2$:
$O\in T_2$ iff one of the following:
1) $O\in T_1$ and $0,1\notin O$
2) $O\in T_1$ and $0,1\in O$
3) $(O\backslash\{0\})\cup\{1\} \in T_1$ and $0\in O,1\notin O$
4) $(O\backslash\{1\})\cup\{0\} \in T_1$ and $0\notin O,1\in O$
This is a topology and the aforementioned sequence converges uniquely to $1$ wrt $T_2$. Is this method right? Are there more elegant examples?
Simple example: consider $X=\mathbb{N}$ and for any $k\in\mathbb{N}$ define $T_k$ to be the smallest topology generated by
$$\text{the interval }[k, \infty)$$ $$\text{almost all singletons }\big\{\{s\}_{s\neq k}\big\}$$
Now consider the sequence $a_n=n$. It can be easily seen that the only two open subsets containing $k$ are $[k,\infty)$ and $X$ and thus $a_n$ converges to $k$ in $T_k$. On the other hand every other point is an open singleton and since $a_n$ is not (eventually) constant then $a_n$ cannot converg to anything other then $k$.