Let $\omega$ be endowed with the discrete topology, and let $\mathbb{R}$ carry the Euclidean topology. Is there a continuous surjective map $f:\omega^\omega\to \mathbb{R}$?


(I suppose this would entail that there is a contiuous surjective map $f:\omega^\omega \to \mathbb{R}^\omega.$)



Is there a continuous surjection $\omega^\omega\to \mathbb{R}$?
Yes - consider the map sending a sequence of naturals to the corresponding continued fraction representation. (OK fine this hits $[0,\infty)$, but that's easy to fix.)


More is true: any [Polish space](https://en.wikipedia.org/wiki/Polish_space) is the continuous image of $\omega^\omega$. The converse fails, though, even for uncountable Polish spaces - there is no continuous surjection from $2^\omega$ to $\omega^\omega$, since only the former is compact.


Cantor space ($2^\omega$) *is* universal in an appropriate sense, though - every compact metric space is a continuous image of Cantor space.


