Need help proving a surjection / uncountability 
Here is what I think.
A) To prove a surjection.. it goes like this. Take an arbitrary b in the Real Numbers (codomain). Let a = "SOMETHING" and we want to show f(a) = b. Now since the function is defined by F(x,y) = x and b is in the real numbers. Everything in the codomain real numbers maps to the domain which is the real numbers. Since f(x,y) = x is in essence the same as f(x) = x. I guess technically each thing is the codomain is mapped to by infinite things in the domain since y can be an infinite number of things.
b) Pretty simple, we can just say since its anything in the real numbers we can count 0.1, 0.11, 0.111, 0.1111.... on and on never getting to one. 
I think I have the gist of each problem down but am having a hard time formalizing it
Thanks for any help!
 A: You intuition for (b) fools you, note that $\mathbb Q$ is countable and all the numbers you listed are rational. Also $f(x,y)$ is not at all the same as $f(x)$ because the domain of functions is important.
Note that $[-1,1]$ is the image of the upper (or the lower) semishpere under $f$
$$G\supset G^+ = \{(x,y)\in G : y\ge 0\}, f(G^+) = [-1,1]$$
and there is an inverse of $f$ on $G^{-1}$ given by $f^{-1}(x) = (x, \sqrt{1-x^2})$.
Depending on your context, proving that $f$ is a surjection is already done by seeing that $(x,\sqrt{1-x^2}) \in G$ is mapped to $x$ under $f$ so any $x\in [-1,1]$ has a preimage.
A: For (a), you should take $b$ in the interval $[-1,1]$ since this is the codomain of $F$.  The reals is not the codomain of $F$.  So if $b \in [-1,1]$, which ordered pair in $G$ is mapping to $b$?  Hint: we know that $(x, y)$ maps to $x$, and also that $(x,y)$ must satisfy $x^{2} + y^{2} = 1$.  So we know a point $(b,\text{something})$ is being mapped to $b$.  How do we find the "something"? Think about what equation that point has to satisfy.  $b$ is a known point.
For (b), here is a better argument:  You need to use the fact from part (a) that you would have already proven by the time you got to part (b).  We know we can find a surjection from $G$ to $[-1,1]$.  What does that mean?  It means everything in $[-1,1]$ is being hit by something in $G$.  But each element of $G$ is going to exactly one place in $[-1,1]$ since functions are well-defined, so no single point can go to more than one place.  But $[-1,1]$ is uncountable, and every element of $[-1,1]$ is being hit by something in $G$, and each element of $G$ is going to exactly one place.  What does this imply about $G$?
