Could you show me how to "calculate" the cardinality of the set of increasing (not necessarily strictly) functions $\ f: \mathbb{Z} \rightarrow \mathbb{Z}$ ?
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1Really, Easy? $x \mapsto x + r$ (defined over $\mathbb{R}$) is real-valued monotonic, and there are as many such maps as there are values of $r$. – Peter Smith May 26 '13 at 13:57
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1[That comment was in reply to someone who asserted that the set of monotonic functions, even over the reals, is countable ...] – Peter Smith May 26 '13 at 14:19
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Similar question with $\mathbb N$ instead of $\mathbb Z$: Cardinality of the set of $ \varphi: \mathbb N \to \mathbb N$ such that $\varphi$ is an increasing sequence – Martin Sleziak May 26 '13 at 17:50
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Hint: For every $A\subseteq\Bbb Z$ you can construct a function which increases exactly at the points of $A$ (i.e. $g(n)<g(n+1)$ if and only if $n\in A$).
(Bonus points: show that if $A\subseteq\Bbb N$ is infinite, then we can use $\{x\in\Bbb Z\mid x\in A\lor -x\in A\}$ to construct a strictly increasing function, conclude that the strictly increasing functions also make a family of size $2^{\aleph_0}$.)
Asaf Karagila
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Let $g\left(n\right)=f\left(n \right)-f\left(n-1\right)$. Then $g:\mathbb{Z}\mapsto\mathbb{Z}_+$ that. by varying $f$, can be any arbitrary sequence of non-negative integers. Knowing $g$ and $f\left(0\right)$ determines $f$, uniquely. So you can look at the cardinality of the set of possible pairs $\left(g,f\left(0\right)\right)$. The cardinality should be the cardinality of $\mathbb{R}$.
S.B.
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Hint:
How many functions are there from $\mathbb{Z}$ to $\mathbb{Z}$? We note that there are $|\mathbb{Z}|$-many values such a function can take at each point $z$ in $\mathbb{Z}$.
How many monotonic functions are there from $\mathbb{Z}$ to $\mathbb{Z}$? We note that there are still $|\mathbb{Z}|$-many values such a function can take at each point $z$ in $\mathbb{Z}$ -- since there are still $|\mathbb{Z}|$-many possible values in the interval $[f(z -1 ), \infty)$ where the value of $f(z)$ must fall.
Peter Smith
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