2

There exists a bijection between $(0,1)$, $(0,1]$ and $[0,1]$?

These 3 sets are not countable and since there are all in $\mathbb{R}$ they should have the same number of elements, so my question which are those bijective functions?

I've been thinking, but I cannot find of any.

Asaf Karagila
  • 393,674
Guadalupe
  • 103
  • You need to use non continuous functions. Enumerate rational numbers and shift them letting irrational numbers fixed. – Crostul Jul 30 '15 at 22:12
  • 1
    Just because they are all super-countable does not mean they are equinumerous. For all you know, there might be cardinalities between that of $\Bbb N$ and that of $\Bbb R$. You're better off making an explicit bijection. – Arthur Jul 30 '15 at 22:13
  • Do they have to be continuous? If so, they don't exist. – Brian Tung Jul 30 '15 at 22:17
  • Note that it suffices for there to be a bijection $f: (0, 1) \to (0, -1]$, for then $g(x) = f(1-x)$ provides a bijection between $(0, 1)$ and $[0, 1)$, and then affixing $1 \to 1$ gives us $(0, 1]$ to $[0, 1]$. – Brian Tung Jul 30 '15 at 22:25

1 Answers1

3

A way to do this is to take a denumerable set in $D \subset (0,1)$, to define a bijection as the identity outside $D$, to map the first element of $D$ to $1$ [and the second to $0$] and to "push" the remaining elements of $D$ down by $1$ [or $2$] to get a bijection with $(0,1]$, and $[0,1]$.

quid
  • 42,135