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.


marked as duplicate by Ross Millikan, Carl Mummert, Mark Bennet, Quixotic, gnometorule Nov 5 '14 at 0:01

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  • $\begingroup$ If you create a bijection, it goes both ways, so you only need one. This has been answered several times on this site. $\endgroup$ – Ross Millikan Nov 4 '14 at 21:26
  • $\begingroup$ 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? $\endgroup$ – Crostul Nov 4 '14 at 21:26
  • $\begingroup$ possible duplicate of How do I define a bijection between $(0,1)$ and $(0,1]$? and this $\endgroup$ – Ross Millikan Nov 4 '14 at 21:28

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

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

  • $\begingroup$ Hotel Hilbert, nice. Hard to believe it's not also an Eagles song. $\endgroup$ – Simon S Nov 4 '14 at 21:57
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    $\begingroup$ @SimonS Never! Not The Eagles. Please! Hilbert's Hotel is too beautiful. But hey, if that's your thing ;) $\endgroup$ – Epsilon Nov 4 '14 at 22:04
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    $\begingroup$ 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. $\endgroup$ – Simon S Nov 4 '14 at 22:07

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


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