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title: 'Summer TIL: Constructing the reals with Dedekind Cuts'

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# Summer TIL: Constructing the reals with Dedekind Cuts

Goal: Construct the real numbers with sets of rational numbers

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Definition of reals:

+ $\mathbb{R}$ is a field under $\times$ and $+$
+ **Totally** ordered with $\leq$
+ $\times$ and $+$ are compatible with the order (relating the ops with the order function)
+ Critical least-upper-bound property: Every non-empty subset bounded above has a least upper bound IN $\mathbb{R}$
    + Not satisfied by $\mathbb{Q}$
    + $\leq$ is "complete"
    + defining property of the reals vs the rationals: no holes, exists supremum

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Great definitions, very intuitive. But how do we know such a set exists?

## Dedekind cuts

Idea: Rational numbers (which we constructed from the naturals) already satisfy most of these properties, so we construct the reals from them and extend their existing operators

Want to fill the "holes" so we use infinite sets of rationals to represent real numbers

Crazy: Define $\mathbb{R}$ as the set of dedekind cuts!

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**define** a cut is a subset $S$ of the rational numbers 

+ $S \neq \emptyset, \neq \mathbb{Q}$
+ No maximum
+ Semi-infinite: if $s \in S, r \in \mathbb{Q}, r<s$, then $r \in S$
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Best (motivating?) property: we get supremum for free because it is just $\bigcup_{s \in S} s$, which is clearly in $\mathbb{R}$!


Let's call rational cuts ($\{x | x \in \mathbb{Q}, x<q\}$) $q^*$

sketch (no proofs but mostly intuitive), just some casework around 0 & signs

+ Addition: $\{a+b | a \in A, b \in B\}$
+ Subtraction: $\{a-b | a \in A, b \in \mathbb{Q} \setminus B\}$
+ Multiplication:
    + Some casework around signs (negate first if not positive), then
    + $\{a \times b | a \geq 0, a \in A, b \geq 0, b \in B\} \cup \{x<0 | x \in \mathbb{Q} \}$
+ Division:
    + Similar, use set of rational divsions by $\mathbb{Q} \setminus B$

+ $\leq$ operator is just $a \subset b$: easy to prove total order!

Need to be careful: ex. traps like defining additive inverse:

Wrong def:
$-A := A + (-A) = 0^*$, as we might have $0 \in -A$ if $A = 0^*$

Cool stuff: extendable to extended real number line by allowing $\emptyset$ or $\mathbb{Q}$ as cuts

### Representing reals as cuts



