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How do I define a bijection between $(0,1)$ and $(0,1]$?

Or any other open and closed intervals?

If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if that's how you're supposed to do it): I make a function $f : (-1, 2)\rightarrow (-5, 4)$ of the form $f(x)=mx+b$ by \begin{align*} -5 = f(-1) &= m(-1)+b \\ 4 = f(2) &= m(2) + b \end{align*} Solving for $m$ and $b$ I find $m=3\text{ and }b=-2$ so then $f(x)=3x-2.$

Then I show that $f$ is a bijection by showing that it is injective and surjective.

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5  
Please don't start your post mid-thought. The subject line is not part of the post, just like the title page is not part of the novel. –  Arturo Magidin Jun 20 '12 at 15:35
    
Try not to double up on your posts. The reason I hadn't gotten back to you after your edit, yet, is that I was sleeping. –  Cameron Buie Jun 20 '12 at 15:56
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"how you're supposed to do it"? This is math! There is no "how you're supposed to do it", there's only "correct" and "incorrect". And, assuming your proofs of injectivity and surjectivity are valid, your construction is correct (one of many correct ways to do it). –  MartianInvader Jun 20 '12 at 23:48

5 Answers 5

up vote 74 down vote accepted

Choose an infinite sequence $(x_n)_{n\geqslant1}$ of distinct elements of $(0,1)$. Let $X=\{x_n\mid n\geqslant1\}$, hence $X\subset(0,1)$. Let $x_0=1$. Define $f(x_n)=x_{n+1}$ for every $n\geqslant0$ and $f(x)=x$ for every $x$ in $(0,1)\setminus X$. Then $f$ is defined on $(0,1]$ and the map $f:(0,1]\to(0,1)$ is bijective.

To sum up, one extracts a copy of $\mathbb N$ from $(0,1)$ and one uses the fact that the map $n\mapsto n+1$ is a bijection between $\mathbb N\cup\{0\}$ and $\mathbb N$.

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7  
+1. This is a beautiful argument. –  user17762 Jun 21 '12 at 6:36
    
Do you know of any sets not made of distinct elements? –  Marc van Leeuwen Aug 31 '12 at 10:01
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@MarcvanLeeuwen This (awkward) formulation is to make sure one understands that $x_n\ne x_k$ for every $n\ne k$. –  Did Aug 31 '12 at 14:05
    
"Choose an infinite sequence $(x_n)_{n\geq1}$ of distinct elements of $(0,1)$" might do. –  Marc van Leeuwen Aug 31 '12 at 14:14
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@GastónBurrull Well, "choose" does not refer to the axiom of choice here but simply to the fact that any such sequence (for example the one you suggest) does the job. (But perhaps your comment was tongue-in-cheek... :-)) –  Did Apr 1 '13 at 8:07

I thought to supplement Did's answer with this picture that I sketched.

enter image description here

The blue is the set $\color{#0073CF}{(0,1) - \{x_n\}^{\infty}_{n \geq 1}}$ and the orange points are elements of the infinite sequence $\color{#FF4F00}{X = \{x_n\}^{\infty}_{n \geq 1}}$. Subscripts (The order of the points) were arbitrarily assigned to each orange points.

Thus, $f : (0,1] \rightarrow (0,1) $ is defined by virtue of:

$f(\color{#FF4F00}{x_n}) = \color{#FF4F00}{x_{n + 1}} \quad \forall \;n \geq 0$ and
$f(\color{#0073CF}{x}) = \color{#0073CF}{x} \qquad \quad \forall \; \color{#0073CF}{x \in {(0,1) - \{x_n\}^{\infty}_{n \geq 1}}}$.

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We will show that both sets are in bijection with $S^1\times \mathbb{Z}$.

Consider $(0,1)$. This is in bijection with $\mathbb{R}$ (for example, scale the interval to $(-\pi/2, \pi/2)$ and apply the tangent function). We can map $\mathbb{R}$ to $S^1\times \mathbb{Z}$ bijectively using the map $t\rightarrow (e^{2\pi i t},\lfloor t \rfloor)$.

Any set homeomorphic to $(0,1]$ can be put into bijection with $S^1$ using the map $t\rightarrow e^{2\pi i t}$. It remains to show that $(0,1]$ is in bijection with countably many copies of itself. To see this, note that the map $x\rightarrow -\frac{1}{x}$ takes $(0,1]$ to $(-\infty, -1]$, and consider the partition

$$\cdots (-4,-3],\ (-3,-2],\ (-2,-1].$$

This seems unnecessarily complicated, and I think you can just map both sets to $\mathbb{R}$ and circumvent the circle stuff, but this is how I figured it out.

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+1. Actually, my mental image of this is a bijection with $(0,1]\times\mathbb Z$ rather than with $S^1\times\mathbb Z$ but this construction is definitely worthwhile to remember. –  Did Jun 21 '12 at 6:56

Let $A=\{\frac{1}{2},\frac{1}{3},...\}$,$B=\{1,\frac{1}{2},\frac{1}{3},...\}$. Define $f:A\rightarrow B$ such that $f(\frac{1}{n})=\frac{1}{n-1}$.It is easy to show that $f$ is a bijection. Then define a function $g:(0,1) \rightarrow (0,1]$ such that

$g(x)=x$ if $x$ is not in $A$ , otherwise $g(x)=f(x)$.

Then $g$ is a required bijection from $(0,1)$ to $(0,1]$.

Remark: We can always solve this kind of question by picking a countable proper subset from (say) $(0,1)$ and then define a bijection $f$ so that the image of $f$ is a little bit bigger than its domain and then define a function which is equal to $f$ on the picked countable set and identity function outside that set.

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1  
thanks that makes it a lot clear for me. If I was to follow that logic for lets say [0,1]^2 and [0,2]^2 would I solve for [0,1] and [0,2] first by letting A = {0,1} and b = {0,2} and then f: A->B ? –  user1411893 Jun 20 '12 at 12:35
    
Between $[0,1]^2$ and $[0,2]^2$, I would rather use the bijection $(x,y)\mapsto(2x,2y)$. –  Did Jun 20 '12 at 14:53

Try something like the function in the following picture:

enter image description here


If you only have to show that such bijection exists, you can use Cantor-Bernstein theorem and $(0,1)\subseteq (0,1] \subseteq (0,2)$. See also open and closed intervals have the same cardinality at PlanetMath.

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