construct a bijection between $(0,1)$ to $R/N$

Question

Suppose $A = (0, 1)$ and $B = R/N$. Then how to construct a bijection from $A$ to $B$? i am aware of bijections from $(0,1)$ to $R$. a hint would be helpful here. where $R$ are real numbers and $N$ are natural numbers and $R/N$ is the set of reals without natural numbers

Probably you want to use either the notation $\mathbb{R}-\mathbb{N}$ or $\mathbb{R}\setminus\mathbb{N}$ to denote the set difference. — Sangchul Lee, Aug 02 '17 at 12:32
I interpreted $R/N$ to mean the quotient, reals mod integers. But now I see he wrote $N$ not $Z$. So logically we do not know what the question is, and have to wait for Nitish to return and tell us. — GEdgar, Aug 02 '17 at 12:49

lasen H · Answer 1 · 2017-08-02T13:15:42.113

$f:x\mapsto\tan \left(\pi \left(x-1/2\right)\right)$ is a bijection from $(0,1)$ to $\mathbb{R}$.

For those $x$ that maps to an integer, you need to redefine $f(x)$ by a non-integer value. You need to choose the new images to be sufficiently fast increasing, because you also need to define those on $x$ that maps to a stolen value.

E.g. something like the redefinitions:
$f(f^{-1}(0))=0.5$
$f(f^{-1}(0.5))=2.5$
$f(f^{-1}(1))=4.5$
$f(f^{-1}(-1))=6.5$
$f(f^{-1}(2))=8.5$
$f(f^{-1}(-2))=10.5$
$f(f^{-1}(2.5))=12.5$
etc...

The resulting function is

$$g(x)=\begin{cases} 2.5 + 5 \left(f(x)-0.5\right) & f(x)\in 2 \mathbb{N}-1.5 \\ 1.5 + 5 f(x) & f(x)\in 2 \mathbb{N}-1\cup -2 \mathbb{N} \\ 0.5 + 5 f(x) & f(x)\in 2 \mathbb{N}_0\cup 1-2 \mathbb{N} \\ f(x) & \text{else} \end{cases}$$

score 0 · Answer 2 · answered Aug 02 '17 at 12:47

We can map from $(0, 1)$ to $\mathbb{R}$:

$$f : (0, 1) \mapsto \mathbb{R} : x \mapsto \frac{2x - 1}{2x(1 - x)}.$$

We can map from $\mathbb{R}$ to $\mathbb{R} \setminus \mathbb{N}$, though not so neatly. Each integer $m$ is the first term of a sequence $(m + 1/n - 1)_{n=1}^\infty$. The trick is to map each sequence into the next term of the sequence. We define a map,

$$g: \mathbb{R} \rightarrow \mathbb{R} \setminus \mathbb{N} : x \mapsto \left\lbrace \begin{array}{cl} m + \frac{1}{n+1} - 1 & \text{if } \exists ~ m,n \in \mathbb{N} : x = m + \frac{1}{n} - 1 \\ 0 & \text{otherwise} \end{array} \right.$$

Then $g \circ f$ is what you need.

score 0 · Answer 3 · answered Aug 02 '17 at 12:48

You can write $$ (0,1)=\left(\frac12,1\right)\cup\bigcup_{n=2}^\infty \left(\frac1{n+1},\frac1 n\right]. $$ Therefore we can consider $(0,1)$ as countable union of disjoints intervals which are open or half open. As mentioned >>here<< you can map half open intervals bijective into open intervalls. You just need to sort which intervall should be mapped to which open intervall in $\mathbb R\setminus \mathbb N$.

construct a bijection between $(0,1)$ to $R/N$

3 Answers3