Prove: Any open interval has the same cardinality of $\Bbb R$ (without using trigonometric functions)

Solution 1:

Consider the function

$$g(x)=\frac{x}{1+|x|}$$ Verify that $g$ is a bijection from real numbers to $(-1,1)$.

Solution 2:

"Uncountable" just means "not countable", where countable is the smallest infinity. If you show that $(a,b)$ is uncountable, and that $\Bbb{R}$ is uncountable, you haven't shown that they have the same cardinality.

You need to exhibit a bijection. That is the very definition of "same cardinality". Any way you prove that the two have the same cardinality will, at least implicitly, exhibit a bijection.

"Same cardinality" means that there is a bijection. So you're asking if you can show that a bijection between these two sets, but without showing that there is a bijection between these two sets. You have to use a bijection. Any theorem, lemma, etc. MUST use bijections, as that is the very definition. What you're asking is like asking "can you show that 2 is even, without showing that 2 is even".

Edit, for an explicit bijection I leave the full construction to you (as it is somewhat tedious), but you could have something like:

$$f(x)=\begin{cases} \frac{1}{x-a} \text{ for } x\in (a,a+\mu/4)\\ \text{ linear connecting term for } x\in [a+\mu/4,a+3\mu/4]\\ \frac{1}{x-b} \text{ for } x\in (a+3\mu/4,b) \end{cases}$$

where $\mu$ is the length of the interval.

Solution 3:

This is an old question, but there is one very simple bijection without trigonometric functions. Consider $f:(0,1) \to \mathbb{R}$ given by

$$f(x) = \begin{cases} \frac{1}{2x} & 0<x<\frac 12 \\ \frac 1{2x-2} + 2, & \frac 12 \leq x < 1 \end{cases}$$

If you want $(a,b)$, it is, of course, simply a matter of scaling/shifting.

Solution 4:

I'll give another bijection, to add to the list (it has a geometric interpretation) :

$x\to \ln(\frac{1}{x-a} -\frac{1}{b-a})$ for $x\in ]a,b[$ is easily seen to be a bijection $]a,b[\to \Bbb{R}$