Cardinal of differentiable functions

Question

Is the cardinal of the set of differentiable functions = card of real numbers? I was thinking of using squeeze rule to show. I know that all differentiable functions are continuous. So card differentiable functions is more than equal to c, but how do I show that its less than or equal to c?

Answer 1

Let $A^B$ denote the set of all functions from a set $B$ to a set $A$. Let $C$ be the set of continuous members of $\mathbb R^{\mathbb R}.$ Let $D$ be the set of differentiable members of $C.$ Let $K$ be the set of constant functions on $\mathbb R.$

If $f$ and $g$ belong to $C$ then $(\forall x\in \mathbb Q\;(f(x)=g(x)))\implies f=g.$

So $\psi (f)=f|_{\mathbb Q}$ is injective from $C$ into $\mathbb R^{\mathbb Q}.$ So $|C|\leq |\mathbb R^{\mathbb Q}|.$ $$ \text {Since }\quad |\mathbb R^{\mathbb Q}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot \aleph_0}=2^{\aleph_0}=|\mathbb R|,$$ $$\text {therefore }\quad |C|\leq |\mathbb R|.$$ Since $K\subset D\subset C ,$ we have $$|\mathbb R|=|K|\leq |D|\leq |C|.$$