# Math Help - Dedekind Cuts?

1. ## Dedekind Cuts?

Given a real number a, define S≡{x|x in Q, x < a}. Prove that a = sup S using dedekind cuts.

Here is a summary of the proof I used:

If ∃ B|B is an upper bound of S and B < a
then (B+a)/2 is not in S.
But (B+a)/2 < a, so we have proof by contradiction.

But I don't know how dedekind cuts would be used in this, since I don't really understand what they are and they aren't used in my text book.

If anyone could point me in the right direction, I would appreciate it.

2. You have asked a very complicated question. The complication results from the wide varieties of notation is use when discussing Dedekind Cuts, Your best bet is find a good text in your library that uses Cuts. A presently popular text is Elementry Analysis by Ken Ross it has a very brief discussion. A very full development can be found in John Randolph’s Basic Real and Abstract Analysis. It has some 15 pages on Cuts but it is an older text.

3. Originally Posted by sjhlax89
Given a real number a, define S≡{x|x in Q, x < a}. Prove that a = sup S using dedekind cuts.

Here is a summary of the proof I used:

If ∃ B|B is an upper bound of S and B < a
then (B+a)/2 is not in S.
But (B+a)/2 < a, so we have proof by contradiction.

But I don't know how dedekind cuts would be used in this, since I don't really understand what they are and they aren't used in my text book.

If anyone could point me in the right direction, I would appreciate it.
Plato is right. (I used Ken Ross' book, it doesn't have much on Dedikind cuts and I never did that in my class either, since it's a stared section and we skipped all of those). I do recall that one of the properties of a Dedikind cut though, is that if a set $S$ is a Dedikind cut and we have $s \in S$, $s \in \mathbb{Q}$. If we have $r \in \mathbb{Q}$ such that $r < s$ then $r \in S$ (or something like that, I don't have my book with me.

If I am correct though, we can do a proof by contradiction using the Denseness of $\mathbb{Q}$ property.

Assume to the contrary that $a \ne \sup S$, but rather, some other element $x = \sup S$. Then, by the definition of the supremum, we must have that $x < a$...

Now apply the Denseness of $\mathbb{Q}$ property to arrive at a contradiction. Can you continue?