# Construct a bijection from $\mathbb{R}$ to $\mathbb{R}\setminus S$, where $S$ is countable

2 inquiries:

Locate a bijective function from $(0,1)$ to $[0,1]$. I have not located the remedy to this given that I saw it a couple of days earlier. It strikes me as weird-- mapping a open set right into a shut set.

$S$ is countable. It's unimportant to locate a bijective function $f:\mathbb{N}\to\mathbb{N}\setminus S$ when $|\mathbb{N}| = |\mathbb{N}\setminus S|$; allow $f(n)$ equivalent the $n^{\text{th}}$ tiniest number in $\mathbb{N}\setminus S$. Exist any kind of similar unimportant remedies to $f:\mathbb{R}\to\mathbb{R}\setminus S$?

A specific bijection $(0,1) \to [0,1]$ for component $1$ is offered by :

$$f\left(\frac{1}{2}\right) = 0,\quad f\left(\frac{1}{3}\right) = 1,\quad f\left(\frac{1}{n}\right) = \frac{1}{n-2}\ \textrm{for}\ n > 3,$$

$$f(x) = x\ \textrm{for}\ x\ \textrm{not equal to a reciprocal of an integer}$$

For a bijection $\mathbb{R}$ to $\mathbb{R}\setminus S$, we can number the components of $S$ (due to the fact that $S$ is countable) as $s_1, s_2, s_3, \dots$ Choose a boundless series $t_1, t_2, \dots$ of distinctive components of $\mathbb{R}$, none of which remain in $S$. After that specify :

$$g(s_i) = t_{2i},\quad g(t_i) = t_{2i+1},\quad g(x) = x\ (\textrm{otherwise}).$$

This is a bijection from $\mathbb{R}$ to $\mathbb{R}\setminus S$.

The evidence of the Schroeder-Bernstein theorem permits you to get a bijection for **1 **, given that we have a shot $(0,1) \to [0,1]$ and also a bijection $f: [0,1] \to [1/4, 3/4] \subset (0,1)$ (claim $x \to x/2 +1/4$). The function's definition will certainly be rather unpleasant (primarily, it relies on the amount of times you can raise a factor under these to shots currently specified, and also especially the parity of the variety of times), yet it'll do it.

For **2 **, iterate this building and construction to get a bijection $R \to R - N$. After that offered any kind of countable set $S$, specify the map of $R$ that swaps $N$ and also $S$ and also leaves every various other factor dealt with. After that the make-up of the first bijection with this 2nd map is your bijection.

Connection factors to consider indicate that the map can not be continual : in **1 **, as an example, we would certainly or else have that $(0,1)$ is portable, which it's not.

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