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


I uploaded my answer to first question and second question.

  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.

  • $\begingroup$ Well it seems it ain't so impossible. By dividing case S$/hat$B is empty and not empty $\endgroup$ – 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 $a<b$.

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,

instead of R

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


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"


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