The "co-small" topology on the naturals?

Consider the set of all subsets of the naturals, $2^\Bbb{N}$. We call a subset $A \subseteq \Bbb{N}$ small if $\sum_{a \in A} \frac{1}{a} < \infty$, and large otherwise. The set of small subsets of $\Bbb{N}$, $$X := \{ A \in 2^\Bbb{N}: A \text{ is small} \},$$ is closed under arbitrary intersection and finite union. This suggests that we can define a topology on $\Bbb{N}$ as follows: Call an element $U \in 2^{\Bbb{N}}$ co-small if $U^c := \Bbb{N} \setminus U$ is small. Then the co-small topology on $\Bbb{N}$ is the topology where the nontrivial open sets (i.e. besides $\Bbb{N}$ and the empty set) are the co-small sets. This is a topology because the set of co-small sets is closed under arbitrary union and finite intersection. Under this topology, $\Bbb{N}$ is $T_1$ (for any two distinct points $a, b$ there is a neighborhood of $a$ disjoint from $b$ and vice versa) but not Hausdorff (since any two co-small sets have co-small intersection, any two neighborhoods of distinct points $a, b$ will overlap). The only compact sets in $\Bbb{N}$ under this topology are the finite sets; however, $\Bbb{N}$ is not discrete in this topology (since one-point sets cannot be co-small).

Questions:

1. Is there a formal name for this topology, and is it studied in the literature at all? Is $\Bbb{N}$ in the co-small topology homeomorphic to another, better known or understood space?

2. What are the continuous functions from $\Bbb{N}$ to itself under the co-small topology, aside from trivial examples like the constant function or the identity function? (The only continuous maps from $\Bbb{N}$ in the co-small topology to $\Bbb{R}$ in the usual topology are the constant functions.)

3. We can also think of $\Bbb{N}$ as a discrete measure space with the obvious measure $$\mu(A) := \sum_{a \in A} \frac{1}{a}.$$ Every open set has infinite measure, and sets with finite measure are closed. Also, every element of $2^\Bbb{N}$ is measurable, from which it trivially follows that every function $f: \Bbb{N} \to \Bbb{R}$ or $\Bbb{C}$ is measurable. The measure also scales in a nice way: we have $\mu(kA) = \frac{\mu(A)}{k},$ where $kA := \{ ka: a \in A \}$. Can this measure be applied to any interesting problems in number theory or combinatorics? Is it used to prove ergodicity of any maps?

Concerning the continuous functions from $\mathbb N$ to itself in the co-small topology:

A map $f\colon \mathbb N\to \mathbb N$ is continuous in the co-small topology if and only if it is either constant or maps large sets to large sets.

I will use the following observations without further notice:

1. Subsets of small sets are small. Super-sets of large sets are large.
2. $\overline X=\mathbb N$ for all large $X$.
3. Co-small sets are large.

Let’s dive into the proof of my claim above. Of course, $f\colon \mathbb N\to \mathbb N$ is continuous if and only if $f(\overline{X})\subset \overline{f(X)}$ for all $X\subset \mathbb N$. If $X$ is closed, this is trivially the case. If $X$ is not closed, i. e., neither small nor all of $\mathbb N$, then $\overline X= \mathbb N$ and so the condition for $f$ to be continuous is that for all large $X$ we have to have $f(\mathbb N)\subset \overline{f(X)}$. There are two essentially different cases to distinguish, namely, whether $f( \mathbb N)$ is small or not.

If $f(\mathbb N)$ is large, then for $f$ to be continuous, $\overline{f(X)}$ must contain a large set, hence, has to be large itself. But then $f(X)$ must have been large to begin with. Conversely, if $f$ maps large sets to large sets, then clearly $f(\mathbb N)\subset \mathbb N=\overline{f(X)}$, as desired.

If $f( \mathbb N)$ is small, then so is each $f(X)$; thus, $f(X)=f(\mathbb N)$ for every large $X$, for $f(X)\subset f(\mathbb N)\subset \overline{f(X)}=f(X)$. I claim that that’s possible only if $f$ is constant.

In fact, let $m\in f(\mathbb N)$ be arbitrary. Since $\{m\}$ is small, it’s closed. Therefore, $f^{-1}m$ is closed, hence either small or all of $\mathbb N$. The latter case means that $f$ is constant. It remains to rule out the former case: If $f^{-1}m$ were small, then $X=\mathbb N-f^{-1}m$ would be co-small, hence large, satisfying $f(X)=f(\mathbb N)-m\subsetneq f(\mathbb N)$, a contradiction. This completes the proof.

I have no idea concerning the 1. and 3. question.