So the title is cryptic, but I was hoping someone could help me understand the following

This assumes a knowledge of Dedekind cuts.

Let us define $0^*$to be the following cut

$0^* = \lbrace p: p<0 \rbrace$

and suppose $\alpha \in \mathbb{R}$, i.e. $\alpha$ is a cut.

Now, presumably, $\alpha \cdot 0^* = 0^* \cdot \alpha = 0^*$

My question is the following: Is this by definition, and if so why are we allowed to simply define it? Or is this something that can be proved, and if so how is it proved?

I'm just not sure how to go about doing this one, and the reference I'm using (Rudin) says virtually nothing about it.

Oh, the other pertinent detail is the way multiplication is defined for Dedekind cuts, which (up the point I've studied) only has done so for positive reals, i.e. give $\alpha, \beta \in \mathbb{R}$, then

$\alpha \beta = \lbrace p: p\leq rs, r\in \alpha, s\in\beta, r>0, s>0 \rbrace$

This is also sometimes defined

$\alpha \beta = \lbrace \lbrace p: p \leq 0\rbrace \cup \lbrace rs : r\in \alpha, s \in \beta, r>0, s>0\rbrace \rbrace$

I.e, all the positive products from alpha and beta and everything below (and including) zero.

Since $0^*$ contains exclusively negative rationals, the above definition for real multiplication can't even be used to discuss the product $\alpha \cdot 0^*$


  • $\begingroup$ What book of Rudin do you use and which chapter? There are several ways to resolve these problems, I think, the simplest is probably to declare $0\cdot a=a\cdot 0=0$ by definition and for non-zero $\alpha$ and $\beta$ multiplication is $\pm|\alpha|\cdot|\beta|$ (for positive numbers it is defined in the question) with the negative sign iff exactly one of $\alpha,\beta$ is negative. $\endgroup$ – Wolfram Feb 7 '17 at 21:10
  • $\begingroup$ Principles of mathematical analysis, Appendix to chapter 1 believe (construction of the reals). Is there really no proof so to speak? It just seems so arbitrary. Dedekind cuts make little enough sense to me as it is, without just declaring things to be true with virtually zero justification. Anyway, pg 29: notendur.hi.is/vae11/%C3%9Eekking/… $\endgroup$ – GeneralPancake Feb 7 '17 at 21:25
  • $\begingroup$ Yes, Rudin defines this as I said. "We confine ourselves first to $\mathbb R^+$, so everything in step $6$ is about positive numbers only. In step 7 he extends multiplication to all $\mathbb R$. Most proofs are omitted there (and in most books) because they are straitforward and tedious. I didn't got about virtually zero justification. There are several constructions of reals, one that is principally different and appeals to me more is via Cauchy sequences, see en.wikipedia.org/wiki/… Dedekind cuts also varies. $\endgroup$ – Wolfram Feb 7 '17 at 21:38
  • $\begingroup$ But the problem is that all constructions lead to some pretty unpleasant proofs. You don't lose much, if you don't understand them now fully though, the constructions of reals require some experience in math from you for not appearing to be black magic. $\endgroup$ – Wolfram Feb 7 '17 at 21:41
  • $\begingroup$ Well so far I've proven everything regarding Dedekind cuts up to multiplication. I want to know the details about the constructions, in case it can answer questions down the road. Cauchy sequences are next. $\endgroup$ – GeneralPancake Feb 7 '17 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.