Proving that cardinality of the reals = cardinality of $[0,1]$ Homework problem, intro to topology. Here's what I've done so far. Am I on the right track? And, how would you advise me to proceed from here?
I have already established that $\left |[0,1] \right | = \left |[0,1) \right|$. Now I wish to show that $\left |[0,1) \right| = \left |\mathbb{R} \right |$. I will try to do this by constructing a bijection $f: [0,1) \rightarrow \mathbb{R}$.
Given $x \in [0,1)$, consider the following countably infinite subset of $\mathbb{R}$: $x^c = \{x+1, x-1, x+2,x-2,x+3,x-3,\ldots   \}$.
Then we have that
$$
\bigcup_{x \in [0,1)} x^c = \mathbb{R},
$$
and if $x_1, x_2 \in [0,1)$ with $x_1 \neq x_2$, then $x_1^c \cap x_2^c = \varnothing$.
 A: $$
\begin{array}{rcl}
[0,1/2) & \longleftrightarrow & (0,1] \\
[1/2,3/4) & \longleftrightarrow & (1,2] \\
[3/4,7/8) & \longleftrightarrow & (2,3] \\
[7/8,15/16) & \longleftrightarrow & (3,4] \\
\vdots & \vdots & \vdots
\end{array}
$$
This gives you a bijection between $[0,1)$ and $(0,\infty)$.  Then $\log$ gives you a bijection between that and $\mathbb R$.
A: Show injections each way and use Cantor-Bernstein theorem to claim a bijection.


*

*$[0, 1]$ injects into $\Bbb R$ as it is.

*$\Bbb R$ injects into $[0, 1]$ as follows
$$f(x)= \begin{cases}
\dfrac1{4x} &\text{for }x > 1&   \text{(injects into }(0, 1/4)\text{ )}\\
\dfrac{1}{4} + \dfrac{x}{4}& \text{for }0\le x\le 1 &\text{(injects into }[1/4, 1/2]\text{ )}\\
\dfrac 12 - \dfrac x4&\text{for }-1<x<0&\text{(injects into }(1/2, 3/4)\text{ )}\\
\dfrac 34 - \dfrac{1}{4x}&\text{for }x\le -1&\text{(injects into }(3/4, 1]\text{ )}\\
\end{cases}
$$
The beauty of CBT is that you don't need to explicitly construct the bijection to claim that it exists and the injections can be anything that works (e.g. no need for continuity, no need to fill the range).  
A: The correspondence $x\leftrightarrow \tan(\pi x + \pi/2)$ is a bijection between $(0,1)$ and $\mathbb R$. If you've already shown that $[0,1]$ and $[0,1)$ have the same cardinality, you should be able to show that $(0,1)$ and $[0,1]$ do, too.
