Prove that the set of lines in the plane is equal in cardinality to $\mathbb{R}$ My attempt:
I begin by trying to find a bijection between $\mathbb{R}^2$ and $\gamma=\{$the set of lines in the plane$\}$. I think the set $\mathbb A = \{ \{(a,-b),(-c,d) \} | \forall a,b,c,d \in \mathbb{R} \}$ would give me all of $\mathbb{R}^2$. I also know that every $x \in \gamma$ is defined by a slope and point. Then there would exists $f$: $\mathbb A \to \gamma$ such that $$f(\mathbb A)=\left ( \dfrac{-b-d}{a+c},(a,-b) \right )$$
The order of the slope and the point obviously do not matter. I am just trying to show that $f$ maps to a point, and slope which uniquely determines each $x \in \gamma$.
If this is correct then $\gamma \equiv_c \mathbb{R}^2 \equiv_c \mathbb{R}$
I see that there are easier ways but is there anything strikingly incorrect about my attempt?
 A: Who says the lines have to be staight or even represent a function in the form y = f(x) ?
Suppose that a line is instead defined parametrically where x = x(t) and y = y(t) and we require that x(t) and y(t) are continuous functions of the real variable t. 
Let X be the set of all such functions {x(t)} and similarly Y = {y(t)}. If L is the set of possible lines then the cardinality |L| = |X x Y| 
But X = Y = C the set of continuous functions of a real variable, and |L| = |C x C|
As a general result, for any set S we have |S x S| = S, so |L| = |C| and the only question is then what is |C|.
A particularly neat result is that a continuous function is compeltely specified by its values at rational points, so that it is equivalent to the set of continuous functions on Q -> Q.
The continuous functions on Q are a subset of all functions Q -> Q, i.e. each continuous function is a subset of Q x Q, so that C is a subset of P(Q x Q) and consequently |C| <= |P(Q x Q)| = |P(Q)| = |R|. I.e |L| <= |R|
It's also easy to see that we can inject R into L by a function f(x) = (x, 1) so that |R| <= |L| 
Putting that together (Cantor Bernstein theorem) then |L| = |R| and there is a bijection between them.
