3

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?

mathmajor
  • 379
  • 1
    Not only is $f$ bijective, but so too is the tangent function. The composition of these in the correct way then results in the bijection you seek. – JMoravitz Nov 17 '17 at 04:06

2 Answers2

7

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:

  1. 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.
  2. 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 $$ |(0,1)| = |\mathbb{R}|. $$

Xander Henderson
  • 29,772
  • I am in an introductory proofs class, and your proof was to the point, and easy to understand. Thank you! – mathmajor Nov 17 '17 at 04:28
1

Follow the hint and use the tangent function. The definition of two sets having equivalent cardinality is that there exists a bijection between them.

spork
  • 11