Seeking a special topology on $\mathbb{N}$

Question

Is there any topology on $\mathbb{N}$ such that every continuous map to a topological space (continuous sequence) is equivalent with convergence of that sequence in the classical sense?? Formally, if $X$ is a topological space, we want a topology on $\mathbb{N}$ such that for every function $f:$ $\mathbb{N}$$\mapsto$$X$, $f$ is continuous if and only if the sequence $f_n$ (the function $f$) is convergent in the classical sense ,(there exists a $x$ on $X$ such that for every neiborhood $U$ of $x$ , there exists a natural number $n_0$ such that for every $n>n_0$ $f_n$ lies on $U$ ). I first thought the topology $T=\{\{n,n+1,n+2,...\} | n\in\mathbb{N}\}$ , but it turns out that every continuous sequence with that topology must be constant, so that topology is inadequate for that purpose.

Answer 1

What you seek is $\omega+1$, in terms of ordinals. It is a sequence with a limit. If you want to stay with just $\Bbb N$, then consider the order topology induced by $\prec_0$ which is defined as $n\prec_00$ for all $n>0$, and $n\prec_0m$ if and only if $n<m$ in the usual order when $n,m\neq 0$.

In other hands, the open sets are any subset of $\Bbb N\setminus\{0\}$ and any co-finite set $A$ such that $0\in A$.

---

Suppose that $f\colon\Bbb N\to X$ is continuous in this topology, then $f(n)\to f(0)$ in $X$. To see why, note that if $U\subseteq X$ is an open neighborhood of $f(0)$, then $f^{-1}(U)$ is open and $0\in f^{-1}(U)$. Therefore $f^{-1}(U)$ contains all but finitely many points, which means that $f(n)\in U$ for all but finitely many $n$'s.