Different limit for two Hausdorff topology? I was working on some exercises and I wonder if the following is true:
Consider a Hausdorff topological space $(X,\tau)$ and a sequence $(x_n)_{n\in \Bbb{N}}$ of elements of $X.$ Suppose that $$x_n\to x\quad\mbox{as}\quad n\to \infty \quad\mbox{for the topology }\tau$$ with $x\in X.$

Now suppose I have another Hausdorff topology on $(X,\tau')$. Assuming that $(x_n)_{n\in \Bbb{N}}$ converges also for this topology, does it follow that$$x_n\to x\quad\mbox{as}\quad n\to \infty \quad\mbox{for the topology }\tau'?$$

All the counter exemples I can found are for non Hausdorff topology, I am convinced that is to "beautiful" to be truth. 
I tried with some probability i.e. convergence in distribution, but we don't really care of $\Omega$ for random variables so...
 A: Consider $[0,1]$. Let $\tau$ denote the usual topology. Define $\tau'$ to be the topology whose neighborhoods are as follows:
For a point $x\in(0,1)$, the basic neighborhoods of $x$ are of the form $(a,b)$, where $0<a<x<b<1$. The basic neighborhoods of $0$ are of the form $\{0\}\cup(a,1)$, where $0<a<1$, and the basic neighborhoods of $1$ are of the form $\{1\}\cup(0,a)$, where $0<a<1$.
Both $\tau$ and $\tau'$ are Hausdorff topologies, and the sequence $(1/n)$ converges to $0$ in $\tau$ and to $1$ in $\tau'$.
A: There is a fundamental question that you have forgotten to ask yourself: how do we compare the sets underlying the two topological spaces?
If you just ask that the underlying sets are in bijection, then consider the following counterexample:

Let $X$ be a set with the cardinality of $\mathbb{R}$, let $(X,\tau)$ and $(X,\tau')$ both be homeomorphic the real line with the standard topology, but such that the roles of $0,1\in(X,\tau)$ are switched in $(X,\tau')$. Then any sequence converging to $0$ in $(X,\tau)$ will converge to $1$ in $(X,\tau')$.

If instead you ask that there is a homeomorphism between the two topological spaces, then the statement is trivially true.
