Prove that $|\mathbb{R}| = |(0, 1)|$. Prove that $|\mathbb{R}| = |(0, 1)|$. (Hint: Consider the tangent function.)
This is my current thought process:

Using the hint, I map $(0, 1) \rightarrow (-\frac{\pi}{2},\frac{\pi}{2})$ by the function $f(x) = \pi x - \frac{\pi}{2}$, and
  then state that since $f$ is linear, and bijective, it must be that -- somehow --
  $|\mathbb{R}| = |(0, 1)|$.

Do I have the general idea? Or am I way off?
 A: Your proof is basically correct, but needs to be fleshed out just a bit.  Recall that two sets have the same cardinality if there is a bijection between them.  We are going to build a bijection from $(0,1)$ to $\mathbb{R}$ in two steps:


*

*Let $\varphi : (0,1) \to (-\frac{\pi}{2},\frac{\pi}{2})$ be the function $\varphi(x) = \pi x - \frac{\pi}{2}$.  As $\varphi$ is linear, it is injective, and it is relatively easy to show that it is surjective on its codomain (continuity plus the intermediate value theorem does the job, though is, perhaps, overpowered for the purpose).  Therefore $\varphi$ is a bijection.

*Let $\psi: (-\frac{\pi}{2},\frac{\pi}{2}) \to \mathbb{R}$ be the tangent function, i.e. $\psi(x) = \tan(x)$.  Since $\tan$ is strictly increasing on this domain, it is injective.  Surjectivity again follows from the intermediate value theorem.  Hence $\psi$ is a bijection.


The composition of bijections is a bijection, and so $\psi\circ\varphi : (0,1) \to \mathbb{R}$ is bijective.  Therefore
$$ |(0,1)| = |\mathbb{R}|. $$
A: Follow the hint and use the tangent function. The definition of two sets having equivalent cardinality is that there exists a bijection between them.
