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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 $f:\omega^\omega \to \mathbb{R}^\omega.$)

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    $\begingroup$ math.stackexchange.com/questions/1225140/… $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 12, 2017 at 13:33
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    $\begingroup$ Just to be clear: if you follow Asaf's link, you'll see that the answer to your question is yes, and that even more is true. You can obtain $\mathbb R$ or $\mathbb R^\omega$ as the continuous bijective image of $\omega^\omega$. In other words, these topologies can be realized as strictly coarser topologies on $\omega^\omega$. (I'm not sure this question will remain open, but personally I don't think it's a bad question -- it's an interesting fact and, though well-known, not trivial to prove if you haven't seen it before.) $\endgroup$
    – Will Brian
    Commented Sep 12, 2017 at 13:38
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    $\begingroup$ I'm voting to close this question as off-topic because it is no longer relevant (answered in the linked question). $\endgroup$
    – Daniel Moskovich
    Commented Sep 12, 2017 at 15:54

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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 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.

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  • $\begingroup$ Brilliant, thanks @NoahSchweber also for the further comments!! $\endgroup$
    – Dominic van der Zypen
    Commented Sep 12, 2017 at 13:59
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    $\begingroup$ Well. Also $\Bbb R$ is connected, so any continuous function to the Baire space is constant. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 12, 2017 at 14:21
  • $\begingroup$ The other one that's often useful is that there is a continuous embedding from any separable metrizable space $X$ to the Hilbert cube $[0,1]^\omega$. (Pick a countable dense subset $\{x_n\} \subset X$ and a metric $d$, and consider the map $x \mapsto (d(x_1, x), d(x_2,x), \dots)$.) $\endgroup$
    – Nate Eldredge
    Commented Sep 12, 2017 at 14:29
  • $\begingroup$ Slight quibble: continued fractions only get you irrational numbers (the map in question is a homeomorphism). There are various simple tricks to get you the rest, see mathoverflow.net/questions/112127/… $\endgroup$
    – Robin Saunders
    Commented Oct 27, 2019 at 23:10
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    $\begingroup$ @RobinSaunders Haha, overlapping comments. I think the state of affairs is: it can make sense, and did to me back when I wrote this post, but current-me was hopelessly confused. $\endgroup$
    – Noah Schweber
    Commented Oct 27, 2019 at 23:58

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