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Order-preserving surjection $\omega^\omega\to${\mathbb N}^{\mathbb N}\to [0,\infty)$

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

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

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

Order-preserving surjection $\omega^\omega\to [0,\infty)$

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

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

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

Order-preserving surjection ${\mathbb N}^{\mathbb N}\to [0,\infty)$

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.

Source Link

Order-preserving surjection $\omega^\omega\to [0,\infty)$

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

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

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