Cardinality the set of strictly decreasing functions $\Bbb N\to \Bbb N$ I'd like some help/clues with finding the cardinality of the set of the strictly decreasing functions from $\Bbb N$ to $\Bbb N$.
I'm not quite sure if it's countable. Hints or clues will be helpful! Thank you so much!
 A: It's zero. There are no strictly decreasing functions $f \colon \mathbf N \to \mathbf N$. Note that for any $n \in \mathbf N$, such an $f$ would have 
$$ f(0) > \cdots > f(n) $$
hence $f(n) \le f(0) - n$. Now let $n = f(0)+1$, this would give 
$$ f\bigl(f(0) + 1\bigr) \le f(0) - f(0) - 1 = -1. $$
Contradiction.
A: Not many strictly decreasing maps $\Bbb N\to \Bbb N$ out there, because a strictly decreasing map $\Bbb N\to\Bbb Z$ satisfies $f(n)\le f(0)-n$
A: There's no such thing as a strictly decreasing function from $\mathbb{N}$ to $\mathbb{N}$.

Thus, the cardinality is $0$.

Now, if you change "strictly decreasing" to "non-increasing" then the set of such functions is countably infinite. To prove it, here's an outline:

$(1)\;$A non-increasing function from N to N must be eventually constant.

$(2)\;$There are a countably infinite number of possibilities for the eventual constant value.

$(3)\;$Fix an eventual constant value, $c$ say. For a given non-increasing function $f:\mathbb{N}\to\mathbb{N}$ with eventual constant value $c$, define the length of $f$ to be the number of terms in the sequence $f(1),f(2),f(3), ...$ which exceed $c$.

$(4)\;$For each $c$ and each length $n$, there are at most countably many non-increasing functions with eventual constant value $c$ and length $n$.

$(5)\;$Finish by noting that a countable union of countable sets is countable.
A: If such sequence exist then :
$\forall n\in \Bbb N \; f(n)\geq 0$
and $f$ strictly decreasing, thus
$\lim_{n\to + \infty}f(n)$ exists $=L$.
so
$f(\Bbb N)=[f(0),L)\cap \Bbb N$ or$[f(0),L]\cap \Bbb N$ which is finite
. This means that the sequence $f(n)$ is stationnary from a certain $n_0$, and this is in contradiction with $f$ is strictly decreasing.
$f$ couldn't exist.
A: martini's answer is correct. You can think of it as a "corollary" to what you should have truly learned from this exercise:
Proposition.
Let $ f:\mathbb{N}\to\mathbb{N} $.
There exists a natural number $ n $ such that $ f\left(n\right)\le f\left(n+1\right) $.
Proof.
Let $ u $ be the least element of $ \operatorname{ran}\left(f\right) $.
Therefore for each $ v\in\operatorname{ran}\left(f\right)\setminus\left\{u\right\} $, $ u<v $.
Because $ u\in\operatorname{ran}\left(f\right) $, there exists $ n\in \mathbb{N} $ such that $ f\left(n\right)=u $.
Because $ n+1\in\mathbb{N} $, $ f\left(n+1\right)\in\operatorname{ran}\left(f\right) $.
Therefore if $ f\left(n+1\right)\ne f\left(n\right) $, then $ f\left(n\right)<f\left(n+1\right) $.
Therefore $ f\left(n\right)\le f\left(n+1\right) $.
