Let us have a general sequence $(a_n)\in\mathbb{N}^\mathbb{N}$ ($a: n\in\mathbb{N}\mapsto a_n\in\mathbb{N}$).
{Question 1}
If we simply insert that sequence into the nested fractions, then we get $$a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\dots}}}\in\mathbb{R}^+$$
It seems that for any such sequence we get a converging sequence, which is a real number. So correct me if I am wrong, but that seems a straightforward bijection of all such sequences onto positive real numbers. Also, we are able to build a ordering simply in the following manner:
def compare(a,b):
i = 0
while (a[i]==b[i]):
if i%2 == 0:
if a[i]<b[i]:
return "a > b"
elif a[i]> b[i]:
return "a < b"
else:
i += 1
else:
if a[i]<b[i]:
return "a < b"
elif a[i]> b[i]:
return "a > b"
else:
i += 1
I never saw that construction alongside others like Dedekind, Weierstrass or Cantor approaches.
{Question 2}
Have anybody investigated the topologies on $\mathbb{N}^\mathbb{N}$ for this mapping to be continous?
UPD:
I found a 1982 publication https://leopard.tu-braunschweig.de/servlets/MCRFileNodeServlet/dbbs_derivate_00031201/Rieger_A_new_approach_to_the_real_numbers.pdf
Rieger, Georg Johann A new approach to the real numbers (motivated by continued fractions) Veröffentlicht in: Abhandlungen der Braunschweigischen Wissenschaftlichen Gesellschaft Band 33, 1982, S.205-217
UPD2:
The main problem of the definition above is $0$, since $1=0+\frac{1}{1}$. So one way or another we need to factorize $\mathbb{N}^\mathbb{N}$.
UPD3:
It seems that the correspondance is more complicated, like $$(\mathbb{Z}×\mathbb{N}^\mathbb{N})\cup (\bigcup_{n\in\mathbb{N}}\bigcup_{(a_k)\in\mathbb{Z}\times\mathbb{N}^n}\{( a_0,a_1,a_2,\dots, a_n,0,0,0,\dots)\}\sim\mathbb{R}$$