Сardinality of the set of increasing functions from Z to Z?

Question

Could you show me how to "calculate" the cardinality of the set of increasing (not necessarily strictly) functions $\ f: \mathbb{Z} \rightarrow \mathbb{Z}$ ?

Answer 1

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 $\mathbb{R}$.

Answer 2

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.

Answer 3

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 $2^{\aleph_0}$.)