Is the set of ALL strictly decreasing functions from $\mathbb{Q }$ to $\mathbb{Q }$ countable? I want to show that the set (call it $B$) of all strictly decreasing functions [$f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that x $<$y implies f(x) > f(y)] is not bijective to the natural numbers N. That is, there exists no bijection  from $B$ to N.
My guess: Since there are not strictly decreasing functions in N (due to positive values), wouldn't that imply that $B$ is not countable? How can I go about from here?
 A: Hint: consider functions of the form 
$$ f(x) = \cases{-x & if $x < r$\cr -1-x & otherwise}$$
where $r \in \mathbb R$.
A: Take any function $g : \mathbb{Z} \to \{0,1\}.$  Then create a strictly decreasing function $f_g : \mathbb{Q} \to \mathbb{Q}$ by defining $$f_g(x) = \begin{cases}g(x)-2x, & \text{ if } x \in \mathbb{Z} \\ (1-\{x\})g(\lfloor x\rfloor) + \{x\}g(\lceil x\rceil) - 2x, & \text{ otherwise}\end{cases}$$
The map $g \mapsto f_g$ is easily seen to be injective.  This shows that $|B| \geq \left|2^\mathbb{Z}\right| = \mathfrak{c}$ is an uncountable cardinal.
A: This is equivalent to proving that there is no enumeration of the set of strictly decreasing functions on $\mathbb Q$. For the sake of contradiction, let $f_n:\mathbb Q\to\mathbb Q$ be a strictly decreasing function for $n\in\mathbb N$ such that $F=\{f_n\}_{n\in\mathbb N}=B$.
We seek to construct a function $g\in B$ such that $g\notin F$. Let $g(1)=f_1(1)-1$. For all $n\in\mathbb N, n>1$, let $$g(n)=\min(g(n-1),f_n(n))-1$$Note that if such a $g$ exists such that $g\in B$, we are done since $g\notin F$. But we know such a $g$ does exist since for all $n\in\mathbb N$, $g(n-1)-g(n)\geq1$, implying that the values the function takes on the naturals is discrete, so we can extend the function to the rationals with no trouble.
This shows that our assumption was invalid, so there exists no such bijection. 
A: Consider a real number $r$. First, construct $f:\mathbb{Z}\to\mathbb{Q}$ strictly decreasing and with $$\lim_{n\to\infty}f(n)=r.$$ It should be easy to prove this is possible.
Now, interpolate linearly, in order to create a strictly decreasing function $f:\mathbb{Q}\to\mathbb{Q}$ with infimum $r$. This creates an injection from $\mathbb{R}$ to $B$, and we’re done.
