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Is there exist a bijective function from $[0, 1)$ to $\mathbb{R}$?

I think it will not possible because $[0, 1)$ is not isomorphic $\mathbb{R}$

Any hints/solution will be appreciated

thanks u

jasmine
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  • Why don't you think it will be possible? Also $[0,1)$ is not isomorphic to $\Bbb{R}$ in what sense? – jgon Nov 06 '18 at 21:13
  • @jgon I believe she means the function cannot be a homeomorphism because it cannot be continuous. – John Douma Nov 06 '18 at 21:18
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    @Sobi Where does $\frac{1}{2}$ go? – John Douma Nov 06 '18 at 21:19
  • @JohnDuoma Ouch, good point! I have removed the comment. – MSDG Nov 06 '18 at 21:20
  • @JohnDouma..can u tell me the particular function ?? – jasmine Nov 06 '18 at 21:21
  • @jasmine I can't off the top of my head. One has to exist because we know that this interval and $\mathbb R$ have the same cardinality. – John Douma Nov 06 '18 at 21:26
  • @JohnDouma I thought that was likely as well, nonetheless my first question was really the more important one. The question is missing context, which is why I posted a comment asking for more information. – jgon Nov 06 '18 at 21:27
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    @jasmine This may be what you need. https://people.math.osu.edu/husen.1/teaching/345/equinumerous_intervals.pdf – John Douma Nov 06 '18 at 21:32
  • @JohnDouma as$ [0,1)$ now removed 0 from $[0,1)$ that will (0,1) which is connected but $ \mathbb{R} -{0}$..not connected as continious image of connected set is connected..here f will not contnious – jasmine Nov 06 '18 at 21:32
  • @JohnDouma....i will read wait for 5 minutes – jasmine Nov 06 '18 at 21:33
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    @jasmine I am aware of that but a bijection does not have to be continuous. – John Douma Nov 06 '18 at 21:33
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    Well, we learned on the streets that the cardinality of $[0,1)$ is $\aleph_1$ and the cardinality of $\mathbb R$ is $\aleph_1$ and we learned in an adult-supervised classroom that the definition of equal cardinality is that a bijection exists. So even if we don't know why we know the answer to whether a bijection exists is .. yes. A bijection must exist because they have the same cardinality. ... That's the answer... now lets try to make sense of it. – fleablood Nov 06 '18 at 21:52
  • @fleablood thanks and Good logics – jasmine Nov 06 '18 at 21:54
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    https://math.stackexchange.com/questions/1425492/explicit-bijection-between-0-1-and-0-1?noredirect=1&lq=1 this might help! – Varazda Nov 06 '18 at 21:59
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    ..And we know we can stretch $0 \to -\infty$ and $1\to \infty$ by... doing something with $\frac 1{something}$ or $\arctan$s or something. So we can probably have $f:(0,1)\to \mathbb R$ if we think about it. But we have an extra point $0$. Where do we put it? Well, we put it to $x_1$ so were do we put $f^{-1}(x_1)$. Well, we put it to $x_2$. Well, where do we put $f^{-1}(x_2)$. Well we put it to $x_3$ and so on. The question is: can we do that? We know the answer is yes. So take a bunch of scratch paper and ... see what you can noodle out. – fleablood Nov 06 '18 at 22:04
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    ..In particular take $0\to f(\frac 12)$ then $\frac 12 \to f(\frac 12)$ then ... $\frac 1{2^k} \to f(\frac 1{2^{k+1}})$. – fleablood Nov 06 '18 at 22:06

2 Answers2

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Start with the identity on $[0,1)$. Then send $0$ to $1/2,$ $1/2$ to $1/3,$ etc. In the end you get a bijection from $[0,1)$ to $(0,1)$. To get a bijection, in fact, a homeomorphism from $(0,1)$ to $\mathbb R$, you can use eg. $\tan(\pi x - \pi /2)$.

You cannot get a continuous bijection of $[0,1)$ to $\mathbb R$ though because then the image of $(0,1)$ has to be an interval. But you get $\mathbb R \setminus\{\text{image of 0}\}.$

jasmine
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Roman Hric
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1

How to get intuition to an answer.

We figure we can find a bijection $f:(0,1)\to \mathbb R = (-\infty, \infty)$ by stretching $0\to -\infty$ and $1 \to\infty$. We aren't sure of the details but we figure we can work those out.

But $[0,1)$ has that $0$ at the endpoint and whatever $f(0)$ maps to then $[0, \epsilon)$ will map to a clopen interval $(f(\epsilon), f(0)]$ or $[f(0),f(\epsilon))$ but we won't be able to continuously map to points immediately to the other side of $f(0)$ and this is because $[0,1)$ is not homeomorphic to $\mathbb R$.

But then we realize the bijection doesn't need to be continuous. And as $(0,1)$ and $[0,1)$ are infinite and $0$ is just one extra point they have a the same cardinality which, by definition, means a bijection must exist.

We can use $f:(0,1) \to \mathbb R$ (once we figure out the details) and then the issue if finding $g:[0,1) \to (0,1)$ that is bijective so $f\circ g:[0,1) \to (0,1) \to \mathbb R$ will be bijective.

We have to map $0\to x_1\ne 0$. That means we must map $x_1\to x_2\ne x_1$ and we must mape $x_2 \to x_3\ne x_2, x_1$. This is basically the Infinity hotel-- make we for the extra point by shoving everyone else one point further. But can we do that in thi interval $(0,1)$? Does there exist an infinite sequence $x_1,x_2, x_3, .....$?

Well, obvious there is. We can do $\frac 12, \frac 13, \frac 14....$ or $\frac 12, \frac 14, \frac 18, ... $ or any we want. So $g:[0,1)\to (0,1)$ can be $g(x) = x$ if $x \not \in \{\frac 1n\}$ and $g(0) = \frac 12$ and $g(\frac 1n) = \frac 1{n+1}$ is a bijection.

(Okay, we'll have to prove this is a bijection but it obviously is. $g:\{\frac 1n; n\ge 2\}\cup\{0\} \to \{\frac 1n; n\ge 2\}$ via $g(\frac 1n)=\frac 1{n+1}$ and $g(0)=\frac 12$ is clearly bijective and $g:(0,1)\setminus \{\frac 1n\}\to (0,1)\setminus \{\frac 1n\}$ via identity is clearly bijective so...)

Now we just need the bijection $f:(0,1)\to \mathbb R$. Well, there is $\arcsin(-\pi 2, \pi 2)\to \mathbb R$ is a bijection and $(0,1) \to (-\pi 2, \pi 2)$ via $x\mapsto \pi(x-\frac 12)$ is a bijection so ... putting it all together:

$h:[0,1)\to \mathbb R$, $h(0) = 0$, $h(\frac 1n; n\ge 2)= \arcsin (\pi*(\frac 1{n+1}-\frac 1n))$ and for all other $x$, $h(x) = \arcsin(\pi (x -\frac 12))$ will be an acceptable bijection.

fleablood
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