Talk:Reverse Triangle Inequality/Real and Complex Fields/Corollary
For the second proof, how can $x \ge y$ imply that $||x|-|y||=|x|-|y|$ ? Take x=1 and y=-2, for example. Then 1 does not equal -1. -- Frades
- Good call. BTW, you can sign your name by clicking on the squiggly signature thing above and to the left.
- Perhaps the proof writer meant: "If $|x| - |y| \ge 0$ or $|x| \ge |y|$, then $||x|-|y|| = |x| - |y|$." --GFauxPas 15:57, 18 March 2012 (EDT)
- Hi folks, thanks for pointing this out-it was my mistaken edit. Of course we needed to have |x|>|y| and |y|>|x| as the cases instead. Everyone should feel free to fix such errors! Noconsciousness 17:10, 18 March 2012 (EDT)
My usual comment.
The scope of this theorem is imprecise. "... or in fact in any normed vector space ..."
As per usual, this is going to need to be stated twice: once as for the most general case possible (which may / may not be "normed vector space") and then the $\R$ and $\C$ cases as specific instances (and proved in the context of real and complex analysis respectively).--prime mover 18:47, 18 March 2012 (EDT)
- Actually, the general form of the theorem is already in Reverse Triangle Inequality. Maybe there should be a link to that page in the "Also See" section? Abcxyz 19:15, 18 March 2012 (EDT)
- Why does this need to be proved separately for $\R$ and $\C$? The current proof, at least, doesn't use any facts about their structure, but only the norm axioms. Relatedly, what imprecision is there in the phrase "normed vector space?" "Norm" has an accepted precise definition. If I'm displaying ignorance of some fundamentals of the ProofWiki community, I apologize and await correction. The main direction of generalization that comes to mind, other than the seminorm, is the metric space, but that's addressed at Abcxyz's link, which I think I'll go ahead and add as suggested. Noconsciousness 19:29, 18 March 2012 (EDT)
- In a nutshell: Users who come looking for help with understanding real analysis will not want to be confused with talk of normed vector spaces. So it is important to provide a proof which relies just on the general mathematical knowledge of real analysis.
- Granted that "normed vector space" is not imprecise, but the general feel of the sentence in which it's worded is flabby to the point of incontinence. --prime mover 19:38, 18 March 2012 (EDT)
- Right-makes sense. Noconsciousness 10:33, 19 March 2012 (EDT)
I suggest to merge this and Reverse Triangle Inequality, using a transclusion setup to distinguish the reals and the metric spaces, because the wording of the titles is atm seriously ambiguous and confusing. --Lord_Farin 11:49, 19 March 2012 (EDT)
- I just thought this again when I revisited this page after a while. --Lord_Farin (talk) 22:48, 26 October 2012 (UTC)
- ... as you see, it is done.
- I expect issue to be taken as to how this page has been set up as a corollary of the other one, when Proof 2 of that result uses this corollary to prove it. But note that I have specifically referenced the precise proof to be used, so as to avoid circularity. Feel free to re-refactor if another way works better for you; I am not completely happy with what I've done here but I can't think of an immediately better way. --prime mover (talk) 07:10, 27 October 2012 (UTC)
- As with every construct where many proofs use a special or simpler case to derive the complete result, there is always some danger of circularity if multiple proofs are added. I think you did a good job on explicitly avoiding it. Any re-refactoring I could imagine would only clutter up the information more and so is undesirable. --Lord_Farin (talk) 07:56, 27 October 2012 (UTC)