Why Can You Negate a Decision but Not a Set & The Problems with {}
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From: Charlie-Boo on 18 Jun 2010 13:22 1. You can negate a decision by switching TRUE and FALSE. For every recursive set its complement is recursive. 2. You can't negate a set because it is really UNIVERSALSET ^ ~SETBEINGNEGATED. You can take a function of two sets P^~Q but there is no ~P in reality. Why the asymmetry? The Problems with {}: 1. You can't really define it. An empty collection? But then why is it a collection? If I define P as containing 0 and also other things, then P is a set because the two place relationship "contains" holds only between sets and things. But {} has no specific elements to make it be a set this way. There is a difference between referring to {0} and {} in that the former is a set just by saying it contains 0 but the latter has no such claim. 2. If you say it contains nothing, then why is a penny not {}? There is nothing that the penny contains. And how do you say that {} is a set, since in (1) we show that you are not really defining what it is. C-B
From: Charlie-Boo on 18 Jun 2010 13:30 On Jun 18, 1:22 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > 1. You can negate a decision by switching TRUE and FALSE. For every > recursive set its complement is recursive. > > 2. You can't negate a set because it is really UNIVERSALSET ^ > ~SETBEINGNEGATED. You can take a function of two sets P^~Q but there > is no ~P in reality. > > Why the asymmetry? > > The Problems with {}: > > 1. You can't really define it. An empty collection? But then why is > it a collection? If I define P as containing 0 and also other things, > then P is a set because the two place relationship "contains" holds > only between sets and things. But {} has no specific elements to make > it be a set this way. There is a difference between referring to {0} > and {} in that the former is a set just by saying it contains 0 but > the latter has no such claim. > > 2. If you say it contains nothing, then why is a penny not {}? There > is nothing that the penny contains. And how do you say that {} is a > set, since in (1) we show that you are not really defining what it is. > > C-B Also, 3. Is {} in PA the same as {} in ZF? In other words, in two systems with different universal sets, is {} the same? If they are different, then how is that difference reflected in the definition? If they are the same, how can the {} in PA handle the question of whether {} contains some particular set when the higher order sets are not able amenable to the processes (functions) of the lower order sets. You can list the universal set in PA but you can't in ZFC. To be more concrete, a computer program to implement PA would not be able to answer the question of whether {0,1,2} contains {}, whereas a program to implement ZFC would.
From: George Greene on 18 Jun 2010 21:12 On Jun 18, 1:22 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > 1. You can negate a decision by switching TRUE and FALSE. For every > recursive set its complement is recursive. However, there are recursively enumerable sets whose complement is NOT also recursively enumerable. These recursively enumerable sets are NOT "just plain" recursive. > 2. You can't negate a set because it is really UNIVERSALSET ^ > ~SETBEINGNEGATED. Oh, shut up. You CAN SO TOO negate a set. Sets have complements. Especially in the context of a FINITE universe. > You can take a function of two sets P^~Q but there is no ~P in reality. If ~P exists as a proposition then it must exist as a set as well. Your personal opinion is not relevant. There are ALREADY DEFINITIONS governing these things! All you are doing is proving that you are either too lazy to read the definitions or too stupid to understand them. > Why the asymmetry? There IS NO asymmetry. You are just alleging one because you don't know any better. In ZFC, it is true that sets don't have "generalized" complements and that there is no universal set. But sets still have "relativized" complements and just because you don't have "the universe" as a set DOES NOT stop you from having sets that are "big enough" for all practical purposes -- nor does it stop you from "having the universe" as a proper class. > > The Problems with {}: None whatsoever. > > 1. You can't really define it. THAT is NEVER a problem! ALL these things are defined BY THE AXIOMS THAT MENTION them (and by other related axioms that use or mention the other things mentioned in the axioms using {} ). This is THE ONLY way to define ANYthing, axiomatically! > An empty collection? But then why is it a collection? BECAUSE *I* *SAID* *SO*, DUH! Why is 1+1=2?? > If I define P as containing 0 and also other things, > then P is a set because the two place relationship "contains" holds > only between sets and things. This is not true. Sets and things ARE NOT different. In ZFC, EVERYthing is a set, so the 2-place relation "is a member of" holds only between SETS and sets of sets. > But {} has no specific elements to make it be a set this way. That's right, so IT is made to be a set IN ANOTHER way: BY DEFINITION.
From: Charlie-Boo on 24 Jun 2010 12:53 On Jun 18, 9:12 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 18, 1:22 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > 1. You can negate a decision by switching TRUE and FALSE. For every > > recursive set its complement is recursive. > > However, there are recursively enumerable sets whose complement is NOT > also recursively enumerable. These recursively enumerable sets are > NOT "just plain" recursive. > > > 2. You can't negate a set because it is really UNIVERSALSET ^ > > ~SETBEINGNEGATED. > > Oh, shut up. > You CAN SO TOO negate a set. > Sets have complements. No, a set plus a universal set has a complement. A set has LOTS of complements. > Especially in the context of a FINITE universe. > > > You can take a function of two sets P^~Q but there is no ~P in reality. > > If ~P exists as a proposition then it must exist as a set as well. Good - even more asymmetry. It does not exist as a single set (see proof above.) > Your personal opinion is not relevant. There are ALREADY DEFINITIONS > governing these things! The definition refers to a universal set as well, so it is not a funciton of a set. > All you are doing is proving that you are either too lazy to read the > definitions or too stupid to understand them. How about too lazy to understand them and too stupid to read them? Stupidity could include illiteracy, you know. > > Why the asymmetry? > > There IS NO asymmetry. You are just alleging one because you don't > know any better. How do you know I'm not doing it to get people riled up (like any smart child)? > In ZFC, it is true that sets don't have "generalized" complements and > that there is no universal > set. The universal set of ZFC is the sets definable by ZFC - a measly aleph-0. > But sets still have "relativized" complements and just because > you don't have "the universe" > as a set DOES NOT stop you from having sets that are "big enough" for > all practical purposes -- This is not practical, this is theoretical. (See the C++ board for practical.) > nor does it stop you from "having the universe" as a proper class. > > > > > The Problems with {}: > > None whatsoever. > > > > > 1. You can't really define it. > > THAT is NEVER a problem! > ALL these things are defined BY THE AXIOMS THAT MENTION them > (and by other related axioms that use or mention the other things > mentioned > in the axioms using {} ). This is THE ONLY way to define ANYthing, > axiomatically! > > > An empty collection? But then why is it a collection? > > BECAUSE *I* *SAID* *SO*, DUH! > Why is 1+1=2?? Lemme see . . . Thm. 1+1=2 1. 1=0' Definition of syntax 1 2. 2=1' Definition of syntax 2 3. 2=0'' Sub 1: 1 => 0' in 2 4. (0')'=0'' Definition of syntax () 5. 0'+0=0' Sub a+0=a (PA): a => 0' 6. (0'+0)'=0'' Sub 5: 0' => 0'+0 in 4 7. 0'+0' = (0'+0)' Sub a+b' = (a+b)' (PA): a=>0' , b=>0 8. (0'+0')=0'' Sub 7: (0'+0)' => 0'+0' in 6 9. 1+1=2 Sub 1: 0' => 1, 3: 0'' => 2 in 8 qed Now who said that it takes 1,000 steps? Those idiots Whitehead and Russell (who said the way to resolve his Russell paradox was to outlaw self-reference!) or the idiot with the MetaMath web site? And they never heard of Occam's Razor?? It takes 1,000 steps if you add unneeded lower level details. But by that reasoning, every proof in mathematics takes 1,000 steps. All could be expanded this way. There is nothing special about 1+1=2 to make it take 1,000 steps. Why should it - it's simple as hell. In fact, most math uses 1+1=2 so all of them need 1,000 steps in their proofs as well. Who could be so stupid as to buy into such silly nonsense??? (Answer: Inbred professors and those with Shared Psychotic Disorders especially those who have bonded emotionally with them. Russell knew all about that as well - young female students who looked up to him as an authoratative figure - his exploits are a well known historical fact.) > > If I define P as containing 0 and also other things, > > then P is a set because the two place relationship "contains" holds > > only between sets and things. > > This is not true. Sets and things ARE NOT different. That wouldn't change anything. Nobody said they had to be different. > In ZFC, EVERYthing is a set, In CBL, you are proven to be an idiot. > so the 2-place relation "is a member of" > holds only between SETS and sets of sets. > > > But {} has no specific elements to make it be a set this way. > > That's right, so IT is made to be a set IN ANOTHER way: BY DEFINITION. Then my penny is the empty set as well. What a joke! C-B
From: William Hale on 26 Jun 2010 03:01 In article <2c95d415-5346-4aa9-855d-5ab83412847a(a)i28g2000yqa.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> wrote: > On Jun 18, 9:12�pm, George Greene <gree...(a)email.unc.edu> wrote: [cut] > > Why is 1+1=2?? > > Lemme see . . . > > Thm. 1+1=2 > 1. 1=0' Definition of syntax 1 > 2. 2=1' Definition of syntax 2 > 3. 2=0'' Sub 1: 1 => 0' in 2 > 4. (0')'=0'' Definition of syntax () > 5. 0'+0=0' Sub a+0=a (PA): a => 0' > 6. (0'+0)'=0'' Sub 5: 0' => 0'+0 in 4 > 7. 0'+0' = (0'+0)' Sub a+b' = (a+b)' (PA): a=>0' , b=>0 > 8. (0'+0')=0'' Sub 7: (0'+0)' => 0'+0' in 6 > 9. 1+1=2 Sub 1: 0' => 1, 3: 0'' => 2 in 8 > qed > > Now who said that it takes 1,000 steps? Those idiots Whitehead and > Russell (who said the way to resolve his Russell paradox was to outlaw > self-reference!) or the idiot with the MetaMath web site? And they > never heard of Occam's Razor?? A similar complaint could be made about Turing Machines. It would be a long program to code up an algorithm to add up two natural numbers. Likewise, using Church's lambda calculus, it would be difficult to code up problems that are routinely given in a beginner's computer programming course. Yet, both of these are of importance. But, I suspect that you might say they are not. > > It takes 1,000 steps if you add unneeded lower level details. Rather, it takes 1,000 steps if you add needed lower level details. That is one of the points. > But by > that reasoning, every proof in mathematics takes 1,000 steps. All > could be expanded this way. > > There is nothing special about 1+1=2 to make it take 1,000 steps. Why > should it - it's simple as hell. In fact, most math uses 1+1=2 so all > of them need 1,000 steps in their proofs as well. If you can give a simple proof from the ZFC axioms that 1 + 1 = 2, nobody would object. But, you gave a simple proof from the Peano axioms. That is like somebody giving the Haskell program "add x y = x + y" when the question was to give a Turing machine program for the "add" function.
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