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I am struggling to get the intuition behind dedekind's cut. The main problem I am facing with Dedekind cut is the third axiom that if (A,B) is dedekind pair for a cut then A should not have largest number. But why does it so?? What I feel that if we know the maximum it can also work as a cut for pair(A,B).

Please send your suggestions. These would be very helpful.


marked as duplicate by Amzoti, Cameron Buie, Alex Wertheim, Jared, Andrey Rekalo Aug 9 '13 at 7:17

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    $\begingroup$ The idea is that if $A$ had a largest element $q\in\Bbb{Q}$, then the pair $(A,B)$ would simply pick out the rational number $q$. So when we're trying to construct irrational real numbers we need pairs observing your axiom. $\endgroup$ – Kevin Carlson Aug 9 '13 at 5:42
  • $\begingroup$ Okay I got your point but i still have one doubt. If we make a cut corresponding to sqrt(2) which means both A and B still have rational numbers. So now why A can't have largest number which is still in Q? $\endgroup$ – Chandresh Sharma Aug 9 '13 at 5:49
  • $\begingroup$ @ Chris yes actually I haven't got answer on that post that's why I have to post it again... $\endgroup$ – Chandresh Sharma Aug 9 '13 at 5:50
  • $\begingroup$ @ChandreshSharma: Because then we would have found $\sqrt{2}\in\mathbb{Q}$, which you can prove is not the case. $\endgroup$ – parsiad Aug 9 '13 at 5:51
  • $\begingroup$ Okay let me put my doubt in much more clear way...lets say i don't have sqrt(2) which truly i don't. So now I accidently have a cut which is actually for sqrt(2), my question is that why the left hand set of sqrt(2) can't have maximum which is very much close to sqrt(2) and still in Q. $\endgroup$ – Chandresh Sharma Aug 9 '13 at 5:53

I think you're confusing two different questions, which have related answers. I'll try to address them both.

Your primary question, judging from the comments, is this: Why must we represent an irrational number $x$ with a cut $(A,B)$ such that $A$ doesn't have a largest number? The answer comes down to what cuts are for. We use cuts to construct a set of numbers that extends the rationals. Every rational number must be represented in the new set of numbers; that's just part of what we're trying to do. Concretely, every rational number must correspond to exactly one cut.

Until we agree upon that goal, it's impossible to continue!

Now, there's a natural way to define the correspondence. For any rational number $q$, we can identify $q$ with either the cut $((-\infty,q],(q,\infty))$ or the cut $((-\infty,q),[q,\infty))$. Both cuts "point to" $q$.

This immediately has two consequences, which neatly answer your questions.

  1. If $(A,B)$ is a cut, and $A$ has a greatest element $q$, then the cut $(A,B)$ represents $q$. In particular, $(A,B)$ is a rational number. Turning this around, if $(A,B)$ is an irrational number then $A$ cannot have a greatest element!
  2. The cuts $((-\infty,q],(q,\infty))$ and $((-\infty,q),[q,\infty))$ are different sets, but they both correspond to the same rational number. In order to ensure that each rational number is represented by only one cut, we must discard one of the two! That is the real purpose of the third axiom. It states that $((-\infty,q],(q,\infty))$ is not a Dedekind cut. As a consequence, $q$ is represented by just one Dedekind cut, namely $((-\infty,q),[q,\infty))$.

Note that we could just as well disallow $((-\infty,q),[q,\infty))$ and accept $((-\infty,q],(q,\infty))$ as a Dedekind cut. Then the modified third axiom would state that $B$ must not have a smallest number. Which convention we choose is merely a matter of taste, so don't worry too much about the choice.

  • $\begingroup$ Thanks Chris it more or less cleared my doubts. $\endgroup$ – Chandresh Sharma Aug 9 '13 at 8:44

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