How do I define a bijection between (0,1) and (0,1]?

[Ganesh Ujwal]

How do I define a bijection between \(\displaystyle (0,1)\) and \(\displaystyle (0,1]\)?


Or any other open and closed intervals?


If the intervals are both open like \(\displaystyle (-1,2)\text{ and }(-5,4)\) I do a cheap trick (don't know if that's how you're supposed to do it):
I make a function \(\displaystyle f : (-1, 2)\rightarrow (-5, 4)\) of the form \(\displaystyle f(x)=mx+b\) by
\(\displaystyle \begin{align*}
-5 = f(-1) &= m(-1)+b \\
4 = f(2) &= m(2) + b
\end{align*}\)

Solving for \(\displaystyle m\) and \(\displaystyle b\) I find \(\displaystyle m=3\text{ and }b=-2\) so then \(\displaystyle f(x)=3x-2.\)


Then I show that \(\displaystyle f\) is a bijection by showing that it is injective and surjective.

 

[pka]

How do I define a bijection between \(\displaystyle (0,1)\) and \(\displaystyle (0,1]\)?
Or any other open and closed intervals?

This is done by 'shifting' a countable set. If \(\displaystyle x=\frac{1}{n}:~n\in\mathbb{Z}^+\) then \(\displaystyle f(x)=\frac{1}{n+1}\); else \(\displaystyle f(x)=x\).

Note that \(\displaystyle f(1)=\frac{1}{2}\) and \(\displaystyle f\left(\frac{1}{2}\right)=\frac{1}{2+1}\).

Now show that \(\displaystyle f:~(0,1] \leftrightarrow (0,1)\).

 

[HallsofIvy]

Equivalently, map every irrational number to itself. The remaining numbers, the rational numbers larger than 0 and less than 1, are countable so can be "listed" {a_i}. Map the first number in that list to 1, then each _i, for I> 1, to a_i+1.