Can convergence be seen as a form of continuity? I will abbreviate "it holds that" to "iht" and "such that" to "sth."
The following questions are motivated by curiosity.
Question 1. Does there exist a topology on $\mathbb{N}$ such that for all topological spaces $Y$ and all sequences $a : \mathbb{N} \rightarrow Y$ iht $a$ is continuous iff $a$ is convergent?
Now suppose we adjoin a maximum element, thereby obtaining $\mathbb{N}' = \mathbb{N} \cup \{\infty\}$.
Question 2. Does there exist a topology on $\mathbb{N}'$ sth for all Hausdorff topological spaces $Y$ and all sequences $b : \mathbb{N}' \rightarrow Y$ iht $b$ is continuous iff the restriction $a : \mathbb{N} \rightarrow Y$ is convergent and has limit equal to $b_\infty$?
Question 3. Suppose the answers are both "yes." Viewing $\mathbb{N}$ as a subset of $\mathbb{N}',$ is the topology induced on $\mathbb{N}$ the same as the topology in Question 1?
 A: 1) No. Suppose such a topology on $\mathbb N$ existed and consider the sequence $a_n=1/n$ as a function $a:\mathbb N \to \mathbb R$. Then $a$ is continuous. Now, consider the same function, but now change the codomain: $a:\mathbb N \to \mathbb R - {0}$. Now $a$ is not continuous. But continuity is a property of the function that is independent of the codomain, so that is impossible. 
2) Yes. Metrize $\mathbb N'$ with $d(m,n)=|1/m-1/n|$ (with the convention that $1/\infty =0$). Then a sequence $a:\mathbb N\to X$, where $X$ is an arbitrary topological space, converges iff it extends to a function $a:\mathbb N'\to X$. If $X$ is Hausdorff, then limits are unique so the condition follows. 
A: The other answers show that 1) doesn't work. Here is some "fix".
Let $O\subset \mathbb N$ be open if and only if $O=\emptyset$ or if $\mathbb N-O$ is finite. For any space $Y$ and $b\in Y$ let $Y_b$ the localisation of $Y$ at $b$, i.e. $Y=Y_b$ as sets and $O\subset Y_b$ open if and only if $O=\emptyset$ or $O\subset Y$ open with $b\in O$.
Then you have
$a:\mathbb N\to Y$ converges to $b$ if and only if $a:\mathbb N\to Y_b$ continuos.
