# Extending a Bijective Map

I have a problem I’m working on that I know the answer to, but I still need some help understanding some things. The problem is as follows:

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

Alright so I know I must provide a bijective function and a possible solution is $tan(\pi(x - \frac{1}{2}))$. This solution makes sense when I look at the graph of the function because it obviously maps the interval $(0,1) \rightarrow \mathbb{R}$ but how would I go about extending this to say the open interval $(a, b)$ where $a < b$.

I’ve been trying to think of a way to stretch and compress the tan function to get a general formula but have failed. Can anyone help me with this problem? Also if you have any other tips for finding bijective maps that would be great. Should I just start to become familiar with as many types of graphs as possible?

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Can you find a bijection between $(0,1)$ and $(a,b)$? – Chris Eagle Feb 2 '13 at 16:48

You should be able to find a simple function of the form $y=mx+c$ that maps $(a,b)$ bijectively to $(0,1)$--$m$ and $c$ will depend on $a$ and $b$, but that's okay, since you're looking for something general, anyway. At that point, just substitute $mx+c$ in for $x$ in $\tan(\pi(x-\frac12))$. The general principle at work, here, is that if $f:A\to B$ and $g:B\to C$ are bijections, then so is $g\circ f:A\to C$.
For another possible bijection $(0,1)\to\Bbb R$, see my answer here (or any of the other fine answers). A similar substitution will allow you to determine a general way to biject $(a,b)\to\Bbb R$.