Finding a bijective function from $\mathbb R \setminus \{0\}$ to $\mathbb R$

Question

Basically I wanna show that $|\mathbb R \setminus \{0\}|=|\mathbb R|$, so I need to find a bijective function $f: \mathbb R \setminus \{0\} \rightarrow \mathbb R$.

What I tried:

$\mathbb R \setminus \{0\} = (-\infty,0) \cup (0, \infty)$

And playing with that. I showed $\mathbb R \setminus \{0\} \sim (-1,0) \cup (0,1) $. What can I do to continue?

Thanks!

Taking $\mathbb N$ to be the nonnegative integers, let $f$ be identity on $\mathbb R\setminus\mathbb N$, and for $x\in\mathbb N$, let $f^{-1}(x)=x+1$. — Lubin, Mar 18 '14 at 13:16
You won't find a continuous function which works, because a bijection between two sets one of which is connected and the other is not cannot be continuous. — Mark Bennet, Mar 18 '14 at 13:23
See also: http://math.stackexchange.com/questions/160738/how-do-i-define-a-bijection-between-0-1-and-0-1 — Martin Sleziak, Mar 18 '14 at 14:47

vadim123 · Accepted Answer · 2014-03-18T13:56:42.523

8

Try this: $$f(x)=\begin{cases} x & x\notin \mathbb{N}\\ x+1 & x\in \mathbb{N}\end{cases}$$

Note: for this function, the domain is $\mathbb{R}$, and $\mathbb{N}=\{0,1,2,3,\ldots\}$. This function is the inverse of the one sought by OP.

edited Mar 18 '14 at 13:56

answered Mar 18 '14 at 13:17

vadim123

82,796

1

And what maps to $1$ under this map? – Mark Bennet Mar 18 '14 at 13:20
@MarkBennet Apparently, $f$ has domain $\Bbb R$ here. – David Mitra Mar 18 '14 at 13:33
@DavidMitra It needs to be as Lubin's comment, with $f(1)=0, f(2)=1, f(3)=2 \dots$ to fill the gap, rather than $f(0)=0, f(1)=2$ which simply shifts the gap. i.e. $f(x)=x-1$. – Mark Bennet Mar 18 '14 at 13:36
@Mark: $f(0)=1$. – Asaf Karagila Mar 18 '14 at 13:38
@AsafKaragila How is $0$ in the domain of the function? The question asks for a function named as $f$ in one direction. This function goes in the other direction, it is true, but to use $f$ which is specified in the question, rather than $f^{-1}(x)$ as in Lubin's comment, or $g(x)$ just causes confusion. – Mark Bennet Mar 18 '14 at 13:48
@Mark: The OP wanted to prove two sets have the same cardinality. Does it really matter which set is the domain and which is the range? – Asaf Karagila Mar 18 '14 at 14:37
@AsafKaragila No. But notational consistency helps. – Mark Bennet Mar 18 '14 at 14:38

score 2 · Answer 2 · edited Mar 18 '14 at 14:49

This is simply a well known generalization of the method proposed by vadim123 which can be used to form a bijection between any two infinite sets where one set contains one more point than the other.

Let $X$ be one infinite set and $X_0 = X \backslash \{y_0\}$ the same set excluding a single point. Take any countable sequence $\{y_k\}$ of elements in $X_0$. Then we define a map

$$f : X \to X_0$$ which essentially pushes elements in $\{y_k\}$ "down the line" while inserting $x_0$ at the position occupied by $y_0$

$$f(x) = \begin{cases}y_0 & \text{if } x = x_0 \\ y_{n+1} & \text{if }x = y_n \\ x & \text{if } x \text{ is not in the sequence } \{y_k\}\end{cases}$$

Side note: Functions of this type are generally proposed as examples of non-continuous maps which which pull back open sets into non-open sets. Take $(0,1)$ into $(0,1]$.

score -4 · Answer 3 · answered Mar 18 '14 at 13:46

-4

Each has the same cardinality. Thus, each is in bijection with the corresponding cardinal number. The composition of bijection is a bijection, so the composition bijects your two sets.

answered Mar 18 '14 at 13:46

8675309

1

I think that the questioner is trying to show that they have the same cardinality, so this is a pretty circular answer. – Simon Rose Mar 18 '14 at 13:53
Not an answer.${}{}$ – Guy Mar 18 '14 at 14:08

Finding a bijective function from $\mathbb R \setminus \{0\}$ to $\mathbb R$

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