How many functions are there?

Claim
There are more real functions than there are real numbers.

Explanation
\(\{f:\mathbb{R}\to\mathbb{R}\} \geq \{f:\mathbb{R}\to\{0, 1\}\} = \mathcal{P}(\mathbb{R}) > \mathbb{R}\)

Claim
There are just as many continuous functions as there are real numbers.

Explanation
\(\{f:\mathbb{R}\to\mathbb{R}, f\text{ continuous}\}\\\qquad\begin{cases}\geq\{f:\{0\}\to\mathbb{R}\} = \mathbb{R}\\\leq\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}=\mathbb{R}^{\mathbb{N}\cdot\mathbb{N}}=2^{\mathbb{N}\cdot\mathbb{N}\cdot\mathbb{N}}=2^\mathbb{N}=\mathbb{R}\end{cases}\)

The first step on the second line is justified by the fact that for every continuous function, there is a sequence of parallelogram functions converging pointwise to it. By parallelogram function, I mean a function whose value is specified at evenly distributed points and interpolated linearly in between. For example, a parallelogram function approximating \(\sin(x)\) is \[f(x) = \sin(\lfloor x\rfloor) + (x-\lfloor x\rfloor)\sin(\lfloor x\rfloor+1).\]
To narrow the number of parallelogram functions down to \(\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}\), require that the value at \(0\) be specified and that the distance between specified points for the \(n\)th function in the sequence be \(1/n\).