Is my proof on showing that the set of monotone functions on $[a,b]$ has cardinality of continum correct? I was given an exercise problem to show that the cardinality of the set of all monotone functions on $[a,b]$ is $\aleph$. I came out with a proof which I am not sure if it is correct. 
My proof: Let $f$ be increasing on $[a,b]$ and $\{x_n\}$ are all discontinuities of $f$. Define 
$$\varphi(x) = \sum_n [f(x_n + 0) - f(x_n)] \theta(x - x_n) + \sum_n[f(x_n) - f(x_n - 0)]\theta_1(x - x_n),$$
where $\theta(x)$ is the Heaviside function and $\theta_1(x) = 1 - \theta(-x)$, and put $g = f - \varphi$, then $\varphi$ is an increasing jump function and $g$ is an increasing continuous function. And we see different monotone functions correspond to different decompositions. On the one hand, $C[a,b]$ has cardinality of continum. On the other hand, we claim that the set of all jump functions on $[a,b]$ also has cardinality $\aleph$, as the process of defining a jump function can be viewed as choosing countably many different points from $[a,b]$ first, and then assigning the corresponding coefficients, so there are $\aleph \cdot \aleph = \aleph$ ways of defining a jump function. Hence by the decomposition, the cardinality of the set of monotone functions cannot exceed $\aleph \cdot \aleph = \aleph$. Since obviously the cardinality is greater than $\aleph$, the proof is completed.
I am not sure whether my argument(the bold part) is correct. I have not studied set theory systematically(All the knowledge I know comes from books on real analysis). Please help me on verifying this proof. Thanks in advance for any help!
 A: Your basic strategy looks like it works, but I think it would be simpler to say something like:
It is clear that there are at least $\mathfrak c$ monotone functions.
On the other hand, every monotone function has at most countably many points of discontinuity, so it is fully specified by the combination of (a) its values at every rational point, (b) the locations of its discontinuities, and (c) its values at every point of discontinuity. So we can specify it using $\omega\cdot 3$ real numbers -- that is, countably many -- and the set of real sequences has cardinality $\mathfrak c$. So there can be at most $\mathfrak c$ monotone functions.
(To unfold such a specification: The specification directly gives the function values at a dense set of points that includes all discontinuities. So the only function values that we need to reconstruct are at points that we know the function is continuous at. For these points we can find the function value as a limit of the points we know; and there are enough of them to find the limit because we know densely many values).

If we're working without the Axiom of Choice there's the additional snag (in either your formulation or mine) that we need to be able to choose a function that takes each monotonic function to a particular one of the many sequences of reals that define it.
For this we need to be able to select a canonical enumeration of the points of discontinuity for all monotone functions at once. But since every discontinuity is a jump discontinuity we can do that. First compress the range of the function into a finite interval, say, by composing it with $y\mapsto e^y/(e^y+1)$; then order the discontinuities in order of decreasing jump height, and for jumps of the same height, in order of increasing $x$.
