This is kind of a continuation of a recent [(closed) question](https://mathoverflow.net/questions/280964/is-there-a-continuous-surjection-omega-omega-to-mathbbr).


Is there an order-preserving surjective function $f:{\mathbb N}^{\mathbb N}\to [0,\infty)$ (where for $a,b\in {\mathbb N}^{\mathbb N}$ we have $a\leq b$ if $a(n) \le b(n)$ for all $n\in {\mathbb N}$)?


Thanks to [Jeremy Rickard](https://mathoverflow.net/users/22989/jeremy-rickard) who made me aware that a previous version of this question was trivial and therefore uninteresting.



Order-preserving surjection ${\mathbb N}^{\mathbb N}\to [0,\infty)$
Yes. Let us isomorphically identify the poset of functions $\omega \to \omega$ (under the pointwise order) with the set of functions $\omega \to \mathbb{N}\_2 = \{n \in \mathbb{N}: n \geq 2\}$, again ordered pointwise. 


Now in fact there is an *isomorphism* of posets $\mathbb{N}\_2^\omega \to [1, \infty)$ given by continued fractions 


$$(a\_1, a\_2, \ldots) \mapsto a\_1 - \frac1{a\_2 - \frac1{a\_3 - \ldots}}$$ 


provided we endow the domain with *dictionary* order. Then finish by observing that the identity function 


$$(\mathbb{N}\_2^\omega)\_{\text{pointwise}} \to (\mathbb{N}\_2^\omega)\_{\text{dict}}$$ 


is order-preserving. 


