# Kinda strange Dedekind cut(partition) as a equivalent of existence of supremum

I was given "this" Dedekind Cut in undergrad. real analysis class

1. A$$\cap$$B=$$\phi$$
2. A$$\cup$$B=R
3. a$$\lt$$b, $$\forall$$ $$a\in A, \forall b\in B$$

$$\to \exists \alpha$$ such that, $$a\le \alpha \le b$$

And I was told this "is" the "Dedekind cut"

But from my memories and from research(not much, mostly wikipedia and here), i think we need

1. if $$a \in A , \exists a' \in$$ A s.t. $$a\lt a'$$

to ensure there exist no largest element in A

Plus, I'm not sure about number 2, their union being set of real number.

Anyway, I had a problem proving "Dedekind cut" $$\Rightarrow$$Existence of Sup(S)

Which definition of $$\alpha$$=Sup(S) given (S: set)

1. $$\alpha \in$$Upperbound(S) (set of upperbounds of S)

2. $$\forall \epsilon \gt$$0, $$\exists x \in$$S s.t. $$\alpha - \epsilon \lt x \leq \alpha$$

(I prefer "if x$$\lt \alpha \to$$ x$$\notin$$Upperbound(S)" though)

if 4. is included in definition, then I have it. (even with not-rigorously given conditions)

1.Are those $$3$$ sufficient for proving existence of Sup($$S$$)?

2.If given "Dedekind cut" was wrong definition and therefore we can't prove existence of Sup($$S$$), then is it possible to prove that it is impossible to prove?

3.Or I just want clear evidence that given definition is wrong in logical way

Edit:

1. So I hope someone can check if it's right.

2. And i wrote there, but I still want to know if it is right to omit "no maximum" statement, and the difference between omitted and not omitted Dedekind cut.

• Well it seems it ain't so impossible. By dividing case S$/hat$B is empty and not empty – tolmekia Mar 15 '19 at 0:42

You have to get some clarity about Dedekind cuts. Instead of dealing with totally ordered sets let's be more specific / concrete and deal with Dedekind cuts of rationals numbers.

A "Dedekind cut" of rationals numbers is a pair of subsets $$A, B$$ of $$\mathbb{Q}$$ such that

• $$A\neq\emptyset \neq B$$
• $$A\cap B=\emptyset$$
• $$A\cup B=\mathbb{Q}$$
• If $$a\in A, b\in B$$ then $$a.

The definition has certain non-obvious consequences in the sense that for any Dedekind cut of rational numbers defined as above we have the following three mutually exclusive and exhaustive possibilities :

• $$A$$ has a greatest member.
• $$B$$ has a least member.
• Neither $$A$$ has a greatest member nor $$B$$ has a least member.

In exactly the same manner one can define a Dedekind cut of real numbers by replacing $$\mathbb{Q}$$ with $$\mathbb {R}$$ in the above definition. But then a surprise awaits us. If sets $$A, B$$ form a Dedekind cut of real numbers then there are two mutually exclusive and exhaustive possibilities :

• $$A$$ has a greatest member.
• $$B$$ has a least member.

This is exactly what is mentioned in the beginning of your post. The proof that there are only two possibilities for a Dedekind cut of reals (compared to three possibilities for a Dedekind cut of rationals) is non-trivial/non-obvious.

Dedekind cuts of rationals are used to define / construct real numbers. When this is done a further condition is added to the definition that the set $$A$$ does not have a greatest member. This is done for technical convenience so that we don't have to deal with three possiblities (out of which first two have similar consequences).

I think I got the answer for first and second question

To begin,

I can bring a set T which is totally ordered set that $$\alpha$$ might or might not be in

1. A$$\cap$$B=$$\phi$$
2. A$$\cup$$B=T T:totally ordered set
3. a$$\lt$$b, $$\forall$$ $$a\in A, \forall b\in B$$

$$\to \exists \alpha \in T'$$ ($$T\subset T'$$) such that, $$a\le \alpha \le b$$ $$\forall a,b \in$$ A,B

Then if $$\alpha \notin$$T then we can just start over by using $$T'$$, for we have formed a new set $$T'$$.

So above saying implies

1. A$$\cap$$B=$$\phi$$
2. A$$\cup$$B=T' T':set formulated from above discussion
3. a$$\lt$$b, $$\forall$$ $$a\in A, \forall b\in B$$

$$\to \exists \alpha \in T'$$ such that, $$a\le \alpha \le b$$ $$\forall a,b \in$$ A,B

Which is same as my given statement.

I think this one is more rigorous and general, for i would not have problem even if T=Q :set of quotient number

so that A$$\cup$$B=Q

Anyway, I'll start proving using T' is total order and algebraic action

S: some set that S$$\subset T'$$

B={b$$\in T$$| $$\forall$$ s $$\in S$$ s$$\leq$$b} set of upperbound

A=B\A

Case 1: If S$$\cap$$B$$\neq \phi$$

than $$\exists \alpha \in$$ S$$\cap$$B. Suppose $$\exists \beta \in$$S$$\cap$$B which $$\alpha \neq \beta$$

by total order $$\beta \lt \alpha$$ and for both in S, which is contradiction that $$\beta$$ cannot be in B

so S$$\cap$$B is singleton {$$\alpha$$}

Thus for $$\alpha \in$$ B, $$\forall x \in S, x \leq \alpha$$ (Upperbound(S)) -1

$$\forall \epsilon \gt$$0, $$\to$$ $$\forall \epsilon$$ $$0\lt \epsilon$$ $$\to$$ $$\forall \epsilon$$ $$\alpha \lt \epsilon +\alpha$$ $$\to$$ $$\forall \epsilon$$ $$\alpha -\epsilon \lt \alpha$$ (basic algebraic action)

$$\forall \epsilon \exists x \alpha -\epsilon \lt x \leq \alpha (x = \alpha$$ would be everytime-candidate of course) -2

by 1,2 such $$\alpha$$ is Sup(S) and thus exists in $$T'$$.

Case 2 If S$$\cap$$B$$= \phi$$

then S$$\subset$$A $$\leftarrow \to \forall x \in S \exists x' \in S s.t. x\lt x'$$

and for x being in A, x$$\leq \alpha$$ (By $$\to \exists \alpha \in T'$$ such that, $$a\le \alpha \le b$$ $$\forall a,b \in$$ A,B)

$$\alpha \in$$ UB(S) -1

Suppose $$\exists \epsilon \gt 0 s.t. \alpha -\epsilon \in B$$. For $$\alpha -\epsilon$$ being one of b $$\in B$$, $$\alpha \leq \alpha -\epsilon$$ which leads to $$\epsilon \leq 0$$ contradiction

Therefore $$\forall \epsilon , \alpha -\epsilon \in A$$

Which implies $$\exists$$ some x that $$\alpha -\epsilon \lt x$$

$$\forall \epsilon \exists x \alpha -\epsilon \lt x \leq \alpha$$ -2

So for first and second question, I think I got answer. (Not sure if it is rigt)

But i still wants to know if that Dedekind cut definition is really a rigorous, educational or right definition.

so I hope someone could give me some difference between Dedekind Cut with no maximum in A and "that Dedekind cut"