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This is kind of a continuation of a recent (closed) question.

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 who made me aware that a previous version of this question was trivial and therefore uninteresting.

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  • $\begingroup$ Just an objection to notation. I think that $\omega$ is an ordinal, and $\omega^\omega$ makes sense in the arithmetic of ordinals. For your set I would write $\mathbb N^\mathbb N$, and then no (sensible) person would think I mean ordinal exponentiation. $\endgroup$
    – Gerald Edgar
    Commented Sep 19, 2017 at 13:28
  • $\begingroup$ Right, thanks @GeraldEdgar - will change $\endgroup$
    – Dominic van der Zypen
    Commented Sep 19, 2017 at 13:30
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    $\begingroup$ @GeraldEdgar I believe that $\omega^\omega$ is relatively common notation for the Baire space. (But I am by no means an expert on descriptive set theory.) Although to be honest, I am a bit confused how this is related to the linked post - which talks about the Baire space. AFAICT only the ordering (not the topology) on the set $\mathbb N^{\mathbb N}$ plays role in this post. $\endgroup$
    – Martin Sleziak
    Commented Sep 19, 2017 at 13:42
  • $\begingroup$ Even if it is common (among set theorists), it is still confusing. Just as confusing as writing $5^2$ for the set of maps from a two-element set to a five-element set. $\endgroup$
    – Gerald Edgar
    Commented Sep 19, 2017 at 13:50
  • $\begingroup$ @Gerald: That last claim seems dubious. This is more like the case of $2^2$. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 19, 2017 at 16:39

2 Answers 2

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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.

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Alternatively, you may map a sequence of positive integers $(a_1,a_2,\dots)$ to $1-2^{-a_1}-2^{-a_1-a_2}-\dots$. This is order-preserving surjection onto $[0,1)$. Apply a function like $\tan (\frac{\pi}2x)$ to get a surjection onto $[0,+\infty)$.

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