Topology defined with convergence

Question

I can understand why two topologies having the same converging sequences doesn't make them equal. But it must make them similar in some sence to be specified, and if so in what possible ways ? Furthermore, I was wondering if two metrics having the same converging sequences makes them topologically equivalent ?

Answer 1

Indeed, two metrics $d_X$ and $d_X'$ on $X$ are topologically equivalent if they have the same converging sequences: let us prove that $\textbf{id} : (X,d_X) \rightarrow (X,d_X')$ is an homeomorphism.

Let $(x_n)_n$ sequence in $X$ such that $x_n \rightarrow x \in X$ for the metric $d_X$. Then $d_X'(\textbf{id}(x_n), \textbf{id}(x)) = d_X'(x_n, x)$ which goes to $0$ as $n \rightarrow \infty$ since both metrics have the same converging sequence, therefore yielding that $\textbf{id}$ is a continuous bijection.

Let us finally prove that $\textbf{id}$ is an homeomorphism by showing that $\textbf{id}^{-1}: (X,d_X') \rightarrow (X,d_X)$ is continuous: the argument is the same as above, thus yielding that $\textbf{id}$ is an homeomorphism.

Therefore, we have that $(X,d_X)$ and $(X,d_X')$ homeomorphic.

Answer 2

There are many types of topologies that cannot be defined in terms of convergent sequences. In particular we can have a topology on a non-empty set $S$ such that $\{p\}$ is not open for any $p\in S,$ and any sequence in $S$ is convergent iff it is eventually constant. (That is, $(p_n)_{n\in N}$ converges to $p$ iff $p_n=p$ for all but finitely many $n.$) There are many of these spaces, and this is all they have in common. A top'l space whose closure operation is definable in terms of convergent sequences is called a sequential space. The order-topology on $\omega_1$ (the first uncountable ordinal) is an example of a sequential space that is not a metrizable space.

Answer 3

Two (metric) topologies with the same converging sequences have the same closed subsets and then the same open subsets, they are therefore equal. Hope it helps.