Convergence and finer topology Can convergent of sequence be used to determine which topology is finer(in general topological space).
I am asking this is question in effect of theorem on metric space: 'topology 1 is finer than topology 2(both on same set) if and only if all cgt sequences in 1 also convergent in 2 to same limit'. and is it true or not if we remove same limit condition from the theorem
 A: Let $X$ be a non-empty set with two metrics $d$ and $e$, and $\tau_d$, $\tau_e$ their induced topologies. You need to show that the following two are equivalent:
$$
d(x_n,x)\to 0 \,\,\Longrightarrow\,\, e(x_n,x)\to 0,
$$ 
and
$$
\tau_e\subset \tau_d.
$$
"$\Longrightarrow$" It suffices to show that if $C\subset X$ is $e-$closed, then it is also $d-$closed. Let $C\subset X$ be $e-$closed and $\{x_n\}_{n\in\mathbb N}\subset C$, with $d(x_n,x)\to 0$. We need to show that $x\in C$. But $d(x_n,x)\to 0$ implies that $e(x_n,x)\to 0$ and hence $x\in C$, as $C$ is $e-$closed.
"$\Longleftarrow$" Assume that $\tau_e\subset \tau_d$ and $d(x_n,x)\to 0$. We need to show that $e(x_n,x)\to 0$. If not there exists an $\varepsilon>0$, such that for every $n_0$, there exists a $n\ge n_0$ satisfying
$$
e(x_n,x)\ge \varepsilon.
$$
In this way we can construct a subsequence $\{x_{k_n}\}$ of $\{x_n\}$ satisfying
$$
e(x_{k_n},x)\ge \varepsilon, \quad \text{for all $n$.}
$$
So
$$
\{x_{k_n}\} \subset \{y\in X: e(x,y)\ge\varepsilon\}=C.
$$
But clearly $C$ is a $e-$closed set and hence, as $\tau_e\subset\tau_d$, a $d-$closed set as well. Thus, the $d-$limit of $\{x_{k_n}\}$ should be in $C$. But the   $d-$limit of $\{x_{k_n}\}$ is $x$, and $x\not\in C$.
A: Yes, indeed. Certain types of topological spaces, such as metric spaces, have the characteristic that the convergence behavior of sequences determines the topology exactly. Such spaces are called sequential spaces.
