# The set of all real functions defined on [0,1] has cardinal number $2^c$

Prove that the set of all real functions defined on the closed unit interval [0,1] has cardinal number $2^c$. it is easy to see that there exists as many as such functions i.e. the characteristic functions but I cannot prove the exact cardinality. so comments will be helpful

-
Any function can be identified with its graph, so the cardinality is at most that of the power set of $[0,1]\times\mathbb R$, i.e. $2^c$ – Etienne Aug 28 '13 at 16:33
Use the fact that $\mathfrak c^\mathfrak c=2^\mathfrak c$ and $[0,1]\sim \Bbb R$. – Git Gud Aug 28 '13 at 16:34
Just below my comment above it says 'Trivial answer converted to comment'. Can anyone else see that? – Git Gud Aug 28 '13 at 16:35
@AsafKaragila Thanks. It disappeared to me too as soon as I posted my question. – Git Gud Aug 28 '13 at 16:36
@SHIBASHIS Then just open the link above. – Git Gud Aug 28 '13 at 16:40

HINT: Use the fact that $|A^B|=|A|^{|B|}$, and show that $|\Bbb{R^R}|=2^{|\Bbb R|}$.

-

To expand on Etienne's comment, we have the following argument:

We have an injection from $\mathcal P([0,1])$ to the set of real functions on $[0,1]$ (which I will call $S$ for convenience) since each subset of $[0,1]$ can be mapped to a unique corresponding characteristic function.

On the other hand, we have an injection from $S$ to $\mathcal P([0,1]\times \mathbb R)$ since each real function in $S$ can be mapped to its unique graph in $[0,1]\times\mathbb R$.

From there, we have only to show that $\left|[0,1]\right|=\left|[0,1]\times\mathbb R\right|$

With the above, we have $$\left|\mathcal P([0,1])\right|\leq \left| S\right| \leq \left|\mathcal P([0,1]\times \mathbb R) \right|=\left|\mathcal P([0,1]) \right|$$ Thus, $\mathcal P([0,1])$ and $S$ have equal cardinalities.

-