Let the set A be defined as follows.

A = { a | real(a) ∧ 0 ≤ a ≤ 1 }

That is, A is the set of real numbers between 0 and 1. Invent a way to represent the members of A using only sets. You are also allowed to use objects that were constructed from sets in the lectures, such as natural numbers and ordered pairs. You must be able to represent a real number with finitely many digits, such as 0.5, and a real number with infinitely many digits, such as π − 3. (These are only examples: you must be able to represent any real number in A.)

I am extremely lost on how to start. I was thinking about summations between 0 and 1, but that didn't get me that far. I am also thinking of how to represent this with an ordered pair.

  • $\begingroup$ en.wikipedia.org/wiki/Dedekind_cut $\endgroup$ – vadim123 Feb 27 '18 at 1:13
  • $\begingroup$ Is there anyway of doing this without using dedekind cuts? $\endgroup$ – user3701380 Feb 27 '18 at 1:14
  • $\begingroup$ en.wikipedia.org/wiki/Cauchy_sequence $\endgroup$ – Noah Schweber Feb 27 '18 at 1:20
  • $\begingroup$ Cauchy sequences can be used to construct the reals from the rationals. $\endgroup$ – Jonathan Feb 27 '18 at 1:21
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    $\begingroup$ This question seems highly dependent on what prior material was in the lectures. Anyway I would represent a real number by a decimal expansion. [edit: This is also mentioned in Lulu's comment from a few seconds before.] $\endgroup$ – Michael Feb 27 '18 at 1:23

Let $S$ be the set of all infinite subsets of $\Bbb N$ and let $T=S\cup \{\emptyset\}.$ For $b\in S$ let $f(b)=\sum_{n\in b}2^{-n}.$ And let $f(\emptyset)=0.$ Then $f:T\to [0,1]$ is a bijection so we can identify each $x\in [0,1]$ with the set $f^{-1}(x).$

For example $f^{-1}(1/3)=\{2n: n\in \Bbb N\}$ and $f^{-1}(1/2)=\Bbb N\setminus \{1\}.$

$f$ is a bijection because for each $x\in (0,1]$ there is a representation $(0. x_1x_2x_3...)$ of $x$ in base-$2$ for which the digit $1$ appears infinitely often, so if $b=\{n:x_n=1\}$ then $f(b)=x.$ And because if $b$ and $b'$ belong to $S$ then $b\ne b'\implies 0\ne f(b)\ne f(b')\ne 0.$


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