Prove $ \left[1,2 \right) \bigcup \left(5,6 \right)$ $=A$ has cardinality? Prove that the following set has cardinality $c$.(which stands for continuum.)
$ \left[1,2 \right) \bigcup \left(5,6 \right)$ $=A$
So I know that A would be a  cardinal number if it is equivalent to$ (0,1).$
So we say 
$f:(0,1) \rightarrow \left[1,2 \right) \bigcup (5,6) $
What I do not understand is how does one derive this function here?
By $   f(x) =
\begin{cases}
2-2x,  & \text{if } 0 < x \le \frac{1}{2} \\
2x+4, & \text{if } \frac{1}{2} < x<1
\end{cases}$
I know that that contiunum is a type of infinum. I just do not know how this function came to be. My hypothesis is that it came from what one substitutes in  for f$(x)$.
 A: To find a bijection from $(0,\frac{1}{2}]$ to $[1,2)$, you can draw the points $(0,2)$ and $(\frac{1}{2},1)$ 
$\;\;$and find the equation of the line joining them.
Similarly, to find a bijection from $(\frac{1}{2},1)$ to $(5,6)$, you can draw the points $(\frac{1}{2},5)$ and $(1,6)$ 
$\;\;$and find the equation of the line joining them.
A: We want $f:(0,1)\rightarrow [1,2) \cup (5,6)$
There's an infinite number of ways to do this.  They chose to find an f: that maps
$f:(0,1/2] \rightarrow [1,2)$ and $f:(1/2, 1)\rightarrow (5,6)$.
And they chose to make this function linear on these two intervals. as linear functions are the simplest bijections.
So they figure  they could do $f(x) = mx + b; x\in (0,1/2]$ and $f(x) = nx + c; x \in (1/2,1)$ where $1 = m*(1/2) + b; 2 = m*2 + b; 5 = n*(1/2)+c; 6=n*(1) + c$.
If we can solve for m,n, b, and c we are done.
A: There is no need of an actual bijection. As @asafKaragila tried to point out in the comment, you just want an injection
$$
f:(0,1) ↪ [1,2) \cup (5,6)
$$
and that is just
$$
f(x) := x + 1
$$
From this, you can see that $c = |(0,1)| ≤ |A|$. On the other side of course $A \subset R$, so $|A| ≤ c$.
Of course, this implies that the two "infinites" are the same.
