Cardinal of an open interval

Question

Lets say there is an open interval (a,b) where a,b are real numbers. Is it true that (a,b) has cardinality equal to real numbers?

Answer 1

This question likely exists on MSE in some form or another; for example, one often encounters a question around finding a bijection between a particular open interval, such as $(0,1)$, and $\mathbb{R}$.

If you check MSE 2203081, then you will find that there is a 1-1 onto map from $(0,1)$ to $\mathbb{R}$ that is defined by $x \mapsto (2x-1)/(x^2 - x)$.

Since you are starting instead with the open interval $(a, b)$, you could use $x \mapsto (x-a)/(b-a)$ in order to create a bijection from $(a, b)$ to $(0, 1)$, and then you could compose this with the function above (or any of the linked examples) that map $(0,1)$ to $\mathbb{R}$ to find a bijection from $(a, b)$ to $\mathbb{R}$.

So: Yes, given any open interval $(a, b)$, it has the same cardinality as the real numbers, as can be demonstrated by a map that puts it into a 1-1 correspondence with $\mathbb{R}$.

Answer 2

Yes. $f(x)=\tan (\frac {\pi}{2}\cdot \frac {x-m}{d}),$ where $m=(b+a)/2$ and $d=(b-a)/2,$ is a bijection from $(a,b)$ to $\mathbb R.$