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How to 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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    Similar question on Quora: https://www.quora.com/What-is-a-bijection-between-0-2-and-0-2?share=12016-02-03

8 Answers 8

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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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    @Parcly That anybody would decide to edit a 5 years old answer, only to replace the (correct) command `\mathbb` by the (deprecated in LaTeX for 20 years) command `\Bbb`, is beyond me.2017-12-06
39

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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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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    Between $[0,1]^2$ and $[0,2]^2$, I would rather use the bijection $(x,y)\mapsto(2x,2y)$.2012-06-20
25

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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    Here is another way to say the same thing. I reverse the half-open intervals. I want to show that the intervals $A=\left[ 0,\infty \right)$ and $B= \left( -\infty,\infty \right)$ are in bijective correspondence. By chopping up in half-open pieces (of length one) like $\ldots, \left[ 3,4 \right) , \left[ 4,5 \right) , \left[ 5,6 \right) ,\ldots$, the interval $A$ is naturally in correspondence $A \approx \left[ 0,1 \right) \times\mathbb{N}$, while for $B$ that is $B \approx \left[ 0,1 \right) \times\mathbb{Z}$. But since it is known that $\mathbb{N}$ and $\mathbb{Z}$ are "equal", we are done.2015-06-30
13

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

enter image description here

The blue line represents the set $\color{#0073CF}{(0,1) - \{x_n\}^{\infty}_{n \geq 1}}$.
The orange circles are elements of the infinite sequence $\color{#FF4F00}{X = \{x_n\}^{\infty}_{n \geq 1}}$, but I plotted only 4 (I chose 4 arbitrarily) circles because it is impossible to plot all elements of an infinite sequence.
Subscripts (The order of the points) were arbitrarily assigned to each orange points.
So the depiction above of $f : (0,1] \rightarrow (0,1) $ can be defined with this formula:

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

2

Clearly, $(\mathbb{R}-\mathbb{Q})\cap [0,1]=(\mathbb{R}-\mathbb{Q})\cap (0,1)$.

So, set some enumeration for the rationals on $[0,1]$, $(r_{n})_{n \ge 1}$, with $r_1 = 0$ and $r_2 = 1$.

Thus, define a function $f:(0,1) \to (0,1]$ to act like the identity on the set of irrationals and, on the set of rationals, set $f(r_j)=r_{j-1}$ for all $j \ge 3$.

This is of course a bijection.

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    This seems to mimick an answer posted four years earlier, only with mistakes which make that the construction does not work. (@upvoter Why the upvote?)2017-04-15
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Note that:

If A is a countable infinite set and a is any element in A, then: there's a bijection for A \ {a} and A;

Or, equivalently if A is contained strictly in B, b is in B \ A, A is countably infinite then: there's a bijection for A $\bigcup$ {b} and A;

To prove the second claim, for example: denote A = {an| n is in $N$}, let: b = f(a0), and an = f(an + 1) for n >= 0. then f is a bijection.

(Similar to a previous answer for your question)

If A is uncountable, you can choose a subset in A as A' that is countable, and do a similar thing to add element to A'. (similar proof for removing element from A, by choosing the element you wish to remove in A')

By this, you can add \ remove as many finite element to an infinite set A as possible, and still have a bijection between them. This is a famous question, called "Hilbert's Hotel", to the name of German Mathematician David Hilbert, you can see more on wiki.

(In fact, (1) if A is countably many, you can even add countably infinite elements to A, and there's still a bijection; and sometimes, you can also do this for removing (2) if A is uncountable, you can always do this for both add/remove a coutably many set to/from A)