cardinalselementary-set-theoryfunctions
I want to prove that every open interval has the same cardinality of $\Bbb R$.
The question is:
Is it enough to prove that any open interval is uncountable? If I prove it, can I say that this interval has the same cardinality of reals?
I know that I can get a bijection using the tangent function, but I am not allowed to use trigonometric functions.
I've proved that $|(a,b)|=|(0,1)|$ so I may prove that $|(0,1)|=|\Bbb R|$ but I did not find the bijection without using trigonometric functions.
Let's do a fundamentally different approach. Consider projecting from the open bottom half of the circle $x^2 + (y-1)^2 = 1$, as I drafted below.
In this, by projecting from $(0,1)$, we get a 1-1 correspondence from the angle $(0, -\pi) \to \mathbb{R}$. There is a clear 1-1 correspondence from $(0,1)$ and $(0, -\pi)$.
Thus we have our bijection from $(0,1)$ to $\mathbb{R}$. For $[0,1]$, we could use big theorems, or we could just modify this one. So let's "make space" by saying that whatever was associated to $0$ is now associated to $2$, what was associated to $1$ is now associated to $3$, $2$ now goes to $4$, and so on, effectively freeing up the real numbers $0,1$ by displacing the natural numbers. Send $0$ to $0$, and $1$ to $1$, and now we have a bijection $[0,1] \to \mathbb{R}$.
Nice, direct, and constructive.
Your first function is a bijection and proves that any two finite intervals have the same cardinality. Your second fails because $g(-\frac 12) \not \in (0,1)$ but you have the right idea. There are many bijections between $\Bbb R$ and some finite interval. One of the simplest is $\arctan(x)\to (\frac {-\pi}2,\frac \pi 2)$ This solves the open intervals. For closed intervals, you need to "swallow" the endpoints somehow.
Best Answer
Consider the function
$$g(x)=\frac{x}{1+|x|}$$ Verify that $g$ is a bijection from real numbers to $(-1,1)$.