I'm stuck on the following question:
Let A be the set of all real-valued functions on [0,1]. show that there does nor exist a function from [0,1] onto A.
I don't even know where to start.
This is a rather advanced theorem.
So I have no way of knowing what you have already proved.
But here are the lemmas you will need.
1) There is an injection $\displaystyle f:X \to Y$ if and only if there is a surjection $\displaystyle g:Y \to X$.
2) There is no surjection $\displaystyle f:X \to P(X)$, (power set of $\displaystyle X$).
3) If $\displaystyle 2^X = \left\{ {f:X \mapsto \{ 0,1\} } \right\}$, the set of functions from $\displaystyle X$ having only two values: 0 or 1.
Then $\displaystyle 2^X \sim P(X)$
Now clearly the set $\displaystyle 2^{[0,1]} \subseteq A = \left\{ {f:[0,1] \to \mathbb{R}} \right\}$
If there were $\displaystyle \varphi :\left[ {0,1} \right]\xrightarrow{{surj}}A$ then
$\displaystyle \varphi :\left[ {0,1} \right]\xrightarrow{{surj}}2^{[0,1]} \sim P\left( {\left[ {0,1} \right]} \right)$.