Is there a easier way to understand dedekind cuts? I tried to understand from Rudin but lost the track after a while. Does anyone have a simpler explanation?
 A: The original definition of Dedekind cut is a bit clumsy... indeed I dont liked at all the first times I approach it. However today I thing they are one of the more beautiful constructions of the mathematics.
The Dedekind cuts can be represented in some different, and simpler, ways than the original one. This is my preferred:
Dedekind cut:
Let define a non-empty subset $R$ of $\Bbb Q$ as a Dedekind cut if it have the following properties:


*

*$R$ dont have a minimum

*If $q\in R$ and $r>q$ then $r\in R$.

*$\Bbb Q\setminus R$ is not empty.
We will call $\Bbb R$ as the set of all the Dedekind cuts. Then we can identify $\Bbb Q$ as a subset of $\Bbb R$ by the injective map
$$\Bbb Q\to\Bbb R,\quad r\mapsto \{x\in\Bbb Q:x>r\}$$
And we define an order relation in $\Bbb R$ by
$$R\supseteq R':\iff R\le R'$$
Suppose that there is some $\mathcal A\subset \Bbb R$ such that is bounded, that is, exists some $B\in\Bbb R$ such that $A\subseteq B$ for all $A\in\mathcal A$, and some $C\in\Bbb R$ such that $C\subseteq A$ for all $A\in\mathcal A$. Then we can define
$$\inf\mathcal A:=\bigcup_{A\in \mathcal A}A,\quad \sup\mathcal A:=\bigcap_{A\in\mathcal A}A\setminus\min\bigcap_{A\in\mathcal A}A$$
Then we can check that these definitions hold for the axiomatic definition of $\Bbb R$.
A: We have the rationals $\Bbb Q$. We want more numbers that fit "between" the rationals. As we don't "have" that numbers yet, we might want to describe them by how they fit between the rationals. That is, for each such number $\alpha$, we should at least be able to tell which rationals are $>\alpha$ and which are $\le\alpha$. Thus for each of our new numbers, we have two subsets $A,B$ of  $\Bbb Q$. And as this relation to the rationals is all we might have got, why not just identify those new numbers with the pair $(A,B)$ of these sets and call them cuts.
What do cuts look like? 


*

*Clearly $A\cap B=\emptyset$ because we want to still have an order relation, i.e., cannot have $q>\alpha$ and $q\le \alpha$ at the same time.

*Clearly $A\cup B=\Bbb Q$ as we want $\alpha$ to be comparable with every rational (and later with every cut, i.e., the order should be total)

*$A\ne \emptyset$ and $B\ne \emptyset$ as we want our new numbers between, not above/below the rationals (well, this might be an afterthought cause if we drop this postulate, we will end up with something like $\Bbb R\cup\{-\infty,+\infty\}$, which is not a field)

*If $x,y\in \Bbb Q$ with $x<y$, then $x\in B$ implies $y\in B$, and $y\in A$ implies $x\in A$; this is just transitivity applied to $\alpha<x<y$ or $x<y\le\alpha$.

*$B$ does not have a minimal element (but $A$ might sometimes have a maximal element)


Lacking ideas for more restrictions, we might just define $\Bbb R$ the the set of all cuts $(A,B)\in\mathcal P(\Bbb Q)\times\mathcal P(\Bbb Q)$ with these five properties.
Actually, I always find it redundant to keep both parts of a cut. Instead, let us work only with the upper part, $B$ (as $A$ is uniquely determined by $B$ and the first two properties). This must have the following properties:


*

*$0\ne B\ne\Bbb Q$

*If $x,y\in\Bbb Q$ with $x<y$, then $x\in\Bbb Q$ implies $y\in \Bbb Q$

*$B$ does not have a minimal element.


So forget about the definition above, we define $\Bbb R$ as the set of subsets of $\Bbb Q$ with these three properties.
Note that for every $q\in \Bbb Q$, the set $\{\,x\in \Bbb Q\mid q<x\,\}$ is an element of $\Bbb R$. This gives us an injective(!) map $\iota\colon \Bbb Q\to\Bbb R$ and via this we sometimes consider $\Bbb Q$ a subset of $\Bbb R$.
We can define an order relation on $\Bbb R$ by declaring $B<B'\iff B'\subsetneq B$. One quickly verifies that this is a total order (i.e., for $\alpha,\beta\in \Bbb R$ exactly one of $\alpha<\beta$, $\beta<\alpha$, $\alpha=\beta$ holds, and $\alpha<\beta\land \beta<\gamma$ implies $\alpha<\gamma$) and it extends the total order of $\Bbb Q$ (i.e., for $a,b\in\Bbb Q$, we have $\iota(a)<\iota(b) \iff a<b$).
We can define addition on $\Bbb R$ by letting $B+B':=\{\,b+b'\mid b\in B,b'\in B'\}$. We readily verify that this is indeed an element of $\Bbb R$. Also $+$ is associative, commutative, and extends the addition on $\Bbb Q$, i.e., $\iota(x)+\iota(y)=\iota(x+y)$ for $x,y\in\Bbb Q$.
To see that $\Bbb R$ is in fact an abelian group under $+$, we need additive inverses. We might be tempted to take $-B:=\Bbb Q\setminus \{\,-x\mid x\in B\,\}$, but this only almost works. This is the first time we encounter a little technicality. The same happens when we later want to define multiplication and have to be careful with signs. But these are just that: technicalities lurking in the details.
In the end, we readily turned teh set $\Bbb R$ of cuts into a field having $\Bbb Q$ (via $\iota$) as a subfield.
A: The idea is to split $\mathbb Q$ into those rationals that are smaller and those that are larger than the real number that the cut is meant to represent. It is enough to use one of these sets.
