1. Proving Equal Cardinality

Problem:

Show that $\displaystyle S = \{x \in \mathbb{R} | 0 < x < 1\}$ has the same cardinality as $\displaystyle \mathbb{R}$.

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From my understanding (and using the book's definition), the cardinality is just the number of elements in the set. So I think I need to prove that there exists a one-to-one function mapping the infinitely many numbers between 0 and 1 to all real numbers $\displaystyle \mathbb{R}$.

I looked at cos(x) and sin(x) to maybe just try to say something like, function f(x) = cos(x) maps all real numbers into a number between 0 and 1, but this is not true. And even if I rigged that to work, it's not a one-to-one function since cos(x), sin(x) etc... can equal the same value with different inputs. Am I on the right track, though?

I don't want the answer, just a little guidance.

Thanks.

2. Re: Proving Equal Cardinality

Originally Posted by tangibleLime
Problem:
Show that $\displaystyle S = \{x \in \mathbb{R} | 0 < x < 1\}$ has the same cardinality as $\displaystyle \mathbb{R}$.
Is the function $\displaystyle \tan\left[\left(-\frac{1}{2}+x\right){\pi}\right]$ one-to-one and onto $\displaystyle S\to\mathbb{R}~?$

3. Re: Proving Equal Cardinality

Thanks for the response! I was working with tan for a bit, but was discouraged about the facts of when it was undefined... I was thinking about it the wrong way.

It seems to me that it would be one-to-one since for each value of x between 0 and 1, it would give a distinct output in $\displaystyle \mathbb{R}$. I think it is also onto since, looking at the graph of this function, it spans the entire codomain off into infinity in both vertical directions.