Correspondence between continuous maps and convergent sequences I am trying to find a topological space $X$ such that there is a bijection between $C(X,Y)$ and $\{convergent$ $sequences$ $in$ $Y\}$ for any Hausdorff space $Y$. The natural idea is to choose $X=\mathbb N$, but then I cannot find a suitable topology. I tried the co-finite topology since it feels like convergent sequence, but with it, any continuous map from $X \to Y$ would be a sequence fluctuating in $Y$ instead of a convergent sequence. Any help would be appreciated. Looking for a explicit construction.
 A: The definition of a convergent sequence is the following. $(x_n)_n$ converges to $x$ iff for each open $U$ around $x$ there exists an $N\in\mathbb{N}$ such that for all $n>N: x_n\in U$. This is the same as saying that the preimage of $U$ is cofinite. But to talk about neighborhoods of the limit $x$ it would be convenient if the limit was also a part of the map in $X\to Y$.
This motivates the bijection between convergent sequences in $Y$ and $C(\mathbb{N}\cup \{\infty\}, Y)$ where the topology on $\mathbb{N} \cup \{\infty\}$ is generated by $\{\{n\}| n\in \mathbb{N}\} \cup \{[n, \infty)\cup\{\infty\}| n\in\mathbb{N}\}$. The important part is that each point in $\mathbb{N}$ is isolated and that each neighborhood of $\infty$ is cofinite.
The bijection sends each convergent sequence to the function $$f:X\to Y: \begin{cases}n\mapsto x_n\\\infty\mapsto \lim_{n\to\infty}x_n\end{cases}$$
The continuity of the function $f$ follows from the initial observation that the preimage of any open $U$ containing the limit is cofinite.*
This bijection is injective, because limits are unique in Hausdorff spaces. And it is also surjective since every continuous function $X\to Y$ gives rise to a convergent sequence by considering the image of $\mathbb{N}\subset X$.
*EDIT:
To explain the continuity of $f$ in more detail, the argument is as follows.
For $f$ to be continuous we need the preimage of any open to be open. So consider any open $V\subset Y$. We split the argument into two cases. Either $V$ contains the limit or $V$ doesn't contain the limit.
Case 1:
$V$ contains the limit, this implies the preimage of $V$ is a cofinite set and the point $\infty$. By definition this set is open.
Case 2:
$V$ does not contain the limit, in this case the preimage of $V$ is a subset of $\mathbb{N}$ because $X$ has the discrete topology on $\mathbb{N}$ all subsets of $\mathbb{N}$ are open, therefore the preimage of $V$ is open.
This proves that the preimage of any open in $Y$ is open in $X$.
