Split the set of real numbers into $2$-element sets

Question

How can we split $\mathbb{R}$ into disjoint sets, each consisting of $2$ elements?

I have found a similar (though much more general) question here.

But I am unable to deduce an answer to my specific question.

When taking the naive approach of $\{(x,-x)|x\in\mathbb{R^+}\}$ we exclude $0$.

Intuitively, I have concluded that $\mathbb{R}$ has an "odd" number of elements.

This leads me to think that splitting $\mathbb{R}$ into $2$-element sets is not feasible.

I do realize, however, that we cannot refer to an infinite number as odd or even.

You can break $\mathbb{Z}$ into $...{-4, -3}, {-2, -1}, {0, 1}, {2, 3}, ...$, and non-integers into ${-x, x}$. — lisyarus, Dec 09 '14 at 14:29
If $x$ is not an integer, partners are $x$ and $-x$, as you suggested. If $x$ is an integer, $x$ and $1-x$ can be partners. — André Nicolas, Dec 09 '14 at 14:30
@lisyarus: $\mathbb{Z}$ is not a problem. In fact, any countable set is not much of a problem, because you can easily "biject" it to $\mathbb{Z}$. My problem is with uncountable sets, such as $\mathbb{R}$. — barak manos, Dec 09 '14 at 14:31
@GrumpyParsnip: OK, I understand that now (from the answer below to be honest). — barak manos, Dec 09 '14 at 14:32

score 3 · Accepted Answer · answered Dec 09 '14 at 14:31

3

First consider the pairs $(2n,2n+1)$ for $n\in\Bbb Z$. Then consider the pairs $(-x,x)$ for $x\notin\Bbb Z$.

answered Dec 09 '14 at 14:31

ajotatxe

65,084

score 2 · Answer 2 · edited May 24 '15 at 10:52

2

How about:

$\{(x,a-x)\mid x\in\mathbb{R^+}, a=1$ if $x\in\mathbb{Z}$ else $a=0\}$

edited May 24 '15 at 10:52

grg

1,017

answered Dec 09 '14 at 14:38

Brian Chandler

358

score 2 · Answer 3 · answered Dec 09 '14 at 15:38

Here's a solution that doesn't treat the integers as special:

Use sets of the form $\{x,x+1 \mid x\in [2n,2n+1), n\in\mathbb{Z}\}$.

The idea is to partition the real line by means of half open unit intervals, and pair corresponding elements of neighboring intervals. So for instance elements of $[-2,-1)$ pair with elements of $[-1,0)$; elements of $[0,1)$ pair with elements of $[1,2)$, etc.

Mikhail Katz · Answer 4 · 2014-12-10T09:45:26.377

0

For negative numbers, just use the pairs $(x,\frac{1}{x})$. For nonnegative numbers, proceed as follows. To make binary representation of real numbers unique, always replace an infinite tail like $0.111\ldots$ by $1.0$, etc. Then pair up numbers that differ only in the first decimal place with respect to this unique representation. Thus, $0.0$ will pair up with $0.1$, while $1.1011\ldots$ will pair up with $1.0011\ldots$, etc.

edited Dec 10 '14 at 09:45

answered Dec 09 '14 at 14:53

Mikhail Katz

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This description is not very clear to me. – barak manos Dec 09 '14 at 14:55

Split the set of real numbers into $2$-element sets

4 Answers4