3

Prove $(0,1)$ and $[0,1]$ have the same cardinality.

I've seen questions similar to this but I'm still having trouble. I know that for $2$ sets to have the same cardinality there must exist a bijection function from one set to the other. I think I can create a bijection function from $(0,1)$ to $[0,1]$, but I'm not sure how the opposite. I'm having trouble creating a function that makes $[0,1]$ to $(0,1)$. Best I can think of would be something like $x \over 2$.

Help would be great.

Extreme112
  • 253
  • If you create a bijection, it goes both ways, so you only need one. This has been answered several times on this site. – Ross Millikan Nov 04 '14 at 21:26
  • If you have a bijection $(0,1) \longrightarrow [0,1]$, then its inverse map is a bijection $[0,1] \longrightarrow (0,1)$. Maybe you meant an injection? – Crostul Nov 04 '14 at 21:26
  • possible duplicate of [How do I define a bijection between $(0,1)$ and $(0,1]$?](http://math.stackexchange.com/questions/160738/how-do-i-define-a-bijection-between-0-1-and-0-1) and this – Ross Millikan Nov 04 '14 at 21:28

2 Answers2

6

Use Hilbert's Hotel.

First identify a countable subset of $(0,1)$, say $H = \{ \frac1n : n \in \mathbb N\}$.

Then define $f:(0,1) \to [0,1]$ so that

$$ \frac12 \mapsto 0$$ $$ \frac13 \mapsto 1$$ $$ \frac{1}{n} \mapsto \frac{1}{n-2}, n \gt 3$$ $$ f(x) = x, \text{for } x \notin H $$

gamma
  • 1,957
  • 1
    Hotel Hilbert, nice. Hard to believe it's not also an Eagles song. – Simon S Nov 04 '14 at 21:57
  • 1
    @SimonS Never! Not The Eagles. Please! Hilbert's Hotel is too beautiful. But hey, if that's your thing ;) – gamma Nov 04 '14 at 22:04
  • 3
    We should give more names to examples or constructions like this. I'm convinced I'll remember this one for some time because of the name together with its elegance. Thanks for posting. – Simon S Nov 04 '14 at 22:07
  • @nwr Is $H$ really a subset of $(0,1)$? After all, $1\in H$ does not fall into $(0,1)$. – Boar Nov 03 '21 at 13:25
  • @Steve Good point. I should have spotted that. – gamma Nov 03 '21 at 20:25
4

Use Cantor-Bernstein theorem:

You can trivially find a bijection between $(0,1)$ and $(1/4,3/4)\subset[0,1]$, hence $\mathrm{Card} (0,1) \leq \mathrm{Card} [0,1]$.

Likewise, there is a trivial bijection between $[1/4,3/4]\subset(0,1)$ and $[0,1]$, hence $\mathrm{Card} [0,1] \leq \mathrm{Card} (0,1)$.

By trivial, I mean a linear function $t\to at+b$ with some numbers $a,b$.

Thus $\mathrm{Card} [0,1] = \mathrm{Card} (0,1)$.

Jean-Claude Arbaut
  • 23,277