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Is there is a bijection from $[0,1)$ to $\mathbb{R}$?

I thought of an instance, $$\frac{\sqrt{x(1-x)}}{x-1}.$$

N. F. Taussig
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Carl
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    Of course there is. They are both sets of size continuum. What makes you think there is not? -1 and vote to close. – Ross Millikan Sep 29 '20 at 02:58
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    @RossMillikan Maybe the OP isn't familiar with cardinality ... – Noah Schweber Sep 29 '20 at 03:00
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    No bijection can be continuous because one has a closed end (0) and the other does not. One point can't spoil anything about bijections with infinite sets because you can always sweep it under the rug. – Ross Millikan Sep 29 '20 at 03:01
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    $\sqrt{x(1-x)}$ is not a bijection from $[0,1)$ to $\mathbb R$. Do you know a bijection from $(0,1)$ to $\mathbb R$? – J. W. Tanner Sep 29 '20 at 03:02
  • If you do know one, then cf. this question – J. W. Tanner Sep 29 '20 at 03:15
  • You don't need to know what a bijection might be to know a bijection must exist. We know a bijection must exist because they have the same cardinality. Is there any reason you care what it is? $x\mapsto \frac {x\sqrt{1-x}}{x-1}$ is not it. (It doesn't map to any negative numbers.) – fleablood Sep 29 '20 at 03:27

3 Answers3

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Define $f:[0,1)=\{0\}\cup\bigcup_{n\ge 1}[\frac1{n+1},\frac1n)\to\mathbb R$ by $f(0)=0$, if $x\in [\frac1{2k+1},\frac1{2k})$, $f(x)=-2k(2k+1)x+3k$, and if $x\in [\frac1{2k+2},\frac1{2k+1})$, $f(x)=(2k+1)(2k+2)x-3k-2$.

A simpler example: $g:[0,1)\to\mathbb R$, if $x\in[0,\frac13]$ then $g(x)=6x-1$, if $x\in (\frac13,\frac23)$ then $g(x)=\frac{1}{2-3x}$, and if $x\in(\frac23,1)$ then $g(x)=\frac{1}{3x-3}$.

Kenta S
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$(0,1) \subset [0,1) \subset \mathbb R$ so $|(0, 1)| \le |[0,1)| \le |\mathbb R|$.

And $f: (0,1)\to (-\frac \pi 2, \frac \pi 2)$ via $x \mapsto \pi x-\frac \pi 2$ is a bijection so $|(0,1)| = |(0,\pi)|$.

And $\tan: (-\frac \pi 2, \frac \pi 2) = \mathbb R$ is a bijection.

So $|\mathbb R|= | (-\frac \pi 2, \frac \pi 2)| = |(0,1)| \le |[0,1)| \le |\mathbb R|$.

So $|[0,1)| = |\mathbb R|$.

So, yes, a bijection between $[0,1)\to \mathbb R$ exist.

But is there any reason we need to actually find one? Isn't enough to to just know there is one?

We can map $(0,1) \to \mathbb R$ by considering $x \mapsto \tan (\pi\cdot x-\frac \pi 2)$. but we haven't got $0$ maping to anything.

We can kludge $[0,1) \to (0,1)$ by mapping $0\mapsto \frac 12$ then $\frac 12 \mapsto \frac 14$ and $\frac 1{2^k} \mapsto \frac 1{2^{k+1}}$ and $x \mapsto x$ if $x \ne \frac 1{2^m}$ for any natural $m$. That is a bijection.

So if we compose the two $[0,1) \to (0,1) \to \mathbb R$ via for every $\frac 1{2^k} \mapsto \tan(\frac 1{2^{k+1}}\pi)$ and for every other $x \mapsto \tan(\pi\cdot x)$.

That's a valid bijection. But there are many others.

It's worth noting a continuous bijection, $f$, can't exist. We need to map $0$ somewhere. But then the values immediately to the left and right of $f(x)$ will be "ripped" apart.

fleablood
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  • $\tan$ is not defined at $\pi/2$. Also, why do you need $1/2^k\mapsto 1/2^{k+1}$? Wouldn't $1/k\mapsto 1/(k+1)$ suffice? – Kenta S Sep 29 '20 at 20:01
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    Isn't $\tan \frac \pi 2$ .... Argh.... brain fart. Use $(0,1)$ to $(-\frac \pi 2, \frac \pi 2)$ instead. As for $\frac 1{2^k} \to \frac 1{2^{k+1}}$ vs. $\frac 1k\to \frac 1{k+1}$. Six of one, half a dozen of the other. Whichever is more intuitive. Which for me is $\frac 1{2^k} \to \frac 1{2^{k+1}}$. – fleablood Sep 29 '20 at 20:13
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Just use usual protocol:

  1. Prove "one to one" (if $f(x)=f(y)$, then $x=y$).
  2. Now we need to show $f$ is onto $\Bbb R$ (This means that $\Bbb R=\text{Range}(f)$).

Now the range of $f$ is the set of values that $f$ maps to from the domain. Now, we need to show that

$$\Bbb R\subset \text{Range}(f)\subset\Bbb R$$

for clear perspective use DESMOS graph calculator you can clearly observe in the graph that given function is not onto.

Souza
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shikamaru nara
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