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-- Which did come first...
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Dj mikey nice sig......its fucking funny

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Originally posted by Chris d(-_-)b I believe some creature close to the chicken genotype evolved to a unstable metaphase. The phenotype of this creature evolved to something close to a chicken. In million years, the genotype was modified and soon it started to resamble a chicken. Now, since the new genotype always had to take characterics through the first stadium of a new individual; the egg, the first real fully evolved chicken was born out of an egg. This leads to our conclusion. The egg wasn't laid by a chicken but rather something close to it. A mutation during the chomosomal duplication of the X or Y chroms lead to the evolvation of the chicken genotype as we know it today. So the egg came first.

I agree with that.

In layman speak it means a thing which was very like a chicken layed an egg which was a bit differnt to that non-chicken (mutated). This thing in the egg was the first thing which could be called a chicken. Which came from an egg and so egg wins.

Yeah?

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Originally posted by Noisician well, if you want to get all technical about that, then the answer really depends on which set theory you base your explanation upon. for example, in zermelo-fraenkel set theory (the one adopted almost universally), the expression "set of all sets" makes no sense whatsoever because in zf such a thing simply does not exist. there's only the CLASS of all sets, and it actually happens to be a proper class (i.e. it's NOT a set). therefore, it cannot be a member of itself because a proper class is NOT an object. this is a direct consequence of the axiom of separation (aka axiom of subsets). for the same reason the entire universe of discourse cannot possibly be a set. it's a proper class as well. and for that matter, we can also prove that the class of all ordinals is not a set. etc.

Nice explanation. I should study some set theory. In any case, it's not interesting for the same reason that Russell's Paradox is.

As for the continued debate for the chicken vs. egg... it's already solved. Depending on how you define an egg (something that gives birth to a chicken or something that a chicken lays), the problem vanishes immediately. If one defines a chicken egg to be "an egg that yields a chicken," then yes, the egg came before the chicken by definition. The silly question "but then what laid THAT egg" becomes moot. Either a chicken laid it, in which case the chicken came from the egg (by defn), or something else laid it, in which case there's still no paradox. If you define it to be "something that a chicken lays," then the chicken came first.

It's a purely logical argument that sidesteps evolution.

Hey I have a question: shut up.

Oh, wait, that wasn't a question.

But still, shut up and stop asking stupid questions that are supposed to make you look really deep but are actually really insipid and nobody really cares about.

If you not interested why post something in this thread? Hmm maybe for a reaction? Feeling lonely?

BTW about

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But still, shut up and stop asking stupid questions that are supposed to make you look really deep but are actually really insipid and nobody really cares about.

Supose you mean look deep and look smart....

...yeah kinda like if you had a pic in your profile which make sure everyone knows you can do FFT calcs isn't it? Or a status which make sure everyone knew you were an EE grad?

....now nobody would care about that would they?

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Originally posted by Dervish If you not interested why post something in this thread? Hmm maybe for a reaction? Feeling lonely? BTW about Supose you mean look deep and look smart.... ...yeah kinda like if you had a pic in your profile which make sure everyone knows you can do FFT calcs isn't it? Or a status which make sure everyone knew you were an EE grad? ....now nobody would care about that would they?

Lol i second that as he invaded my thread, twat.

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Originally posted by Dervish If you not interested why post something in this thread? Hmm maybe for a reaction? Feeling lonely? BTW about Supose you mean look deep and look smart.... ...yeah kinda like if you had a pic in your profile which make sure everyone knows you can do FFT calcs isn't it? Or a status which make sure everyone knew you were an EE grad? ....now nobody would care about that would they?

Damn, you caught me, and here I was thinking I was going to get away with it. A lot of people said they thought the avatar and status were funny, you know, due to there being no jobs for EEs anymore, but I guess they were just humouring me and trying not to hurt my feelings. I'm just starved for attention and the only place I can get it is online, I just wish you hadn't found me out!

...or, maybe I just thought it was a stupid thread. Occam's razor, anyone?

rules of evolution say egg.

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Originally posted by DigiNut Damn, you caught me, and here I was thinking I was going to get away with it.

Away with what?.... being a wanker?

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due to there being no jobs for EEs anymore,

... maybe it's just you.

Besides lets have your opinion chicken or egg? If you don't provide one then I don't see the point in you answering.

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Originally posted by DigiNut But still, shut up and stop asking stupid questions that are supposed to make you look really deep but are actually really insipid and nobody really cares about.

I don't think that's what he's going for.

I think he's just your token asinine 13 year old.

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Originally posted by Flyboy217 Nice explanation. I should study some set theory. In any case, it's not interesting for the same reason that Russell's Paradox is.

you don't think russell's paradox is interesting enough? hmm... i think it's amazing how one little antinomy followed from apparently sound assumptions was able to plunge the entire set theory into a crisis at the time it was devised. the paradox destroyed one of the most plausible postulates that cantor and his adherents piously believed in. what russell did was checkmate cantor's comprehension principle in two simple moves. unlike previous antinomies that had arisen before, russel's one did not require lengthy expositions and seemed very sound even to persons completely unfamiliar with set theory. he actually showed that some classes could not be regarded as sets. sure cantor had already been partially aware of this contradiction (he knew of the burali-forti paradox) but he didn't devote attention to this discovery at all, nor was he any disturbed by it. and boy, was he wrong. this clever trick of russel's persuaded zermelo into developing what later became known as zermelo-fraenkel set theory, the one we still use nowadays. hell, it still remains the best we have. though i should also add that the axiomatic approach it's based on is not ideal, either. while it helps with blocking the logical antinomies, it does not guard against various semantic antinomies, such as berry's paradox for example.

oh and btw, a chicken is just the means by which one egg produces another egg.

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Originally posted by Noisician you don't think russell's paradox is interesting enough? hmm... i think it's amazing how one little antinomy followed from apparently sound assumptions was able to plunge the entire set theory into a crisis at the time it was devised. the paradox destroyed one of the most plausible postulates that cantor and his adherents piously believed in. what russell did was checkmate cantor's comprehension principle in two simple moves. unlike previous antinomies that had arisen before, russel's one did not require lengthy expositions and seemed very sound even to persons completely unfamiliar with set theory. he actually showed that some classes could not be regarded as sets. sure cantor had already been partially aware of this contradiction (he knew of the burali-forti paradox) but he didn't devote attention to this discovery at all, nor was he any disturbed by it. and boy, was he wrong. this clever trick of russel's persuaded zermelo into developing what later became known as zermelo-fraenkel set theory, the one we still use nowadays. hell, it still remains the best we have. though i should also add that the axiomatic approach it's based on is not ideal, either. while it helps with blocking the logical antinomies, it does not guard against various semantic antinomies, such as berry's paradox for example. oh and btw, a chicken is just the means by which one egg produces another egg.

No no sorry, you misunderstand. I said the question "does the set of all sets contain itself" is not interesting in the way that Russell's Paradox is. Clearly Russell's Paradox is *very* interesting. Godel perhaps wouldn't have been motivated to produce On Principia Mathematica if not for this. And then where would we be...

stupidest shit I've ever heard

P.S. God doesn't exist

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Originally posted by Flyboy217 No no sorry, you misunderstand. I said the question "does the set of all sets contain itself" is not interesting in the way that Russell's Paradox is.

all right, then

but i still think the issue he brought up is engaging AND important. the thing is, the deeper you look into it, the less basic it appears to be. admittedly, one tends to intuitively assume that the set of all sets would surely contain itself (after all, it includes ALL sets, right?), but it actually turns out that such a structure cannot possibly exist in the universe of discourse. it's not a valid object in the sense that it is not subject to the fundamental axioms of modern set theory. as a result, it cannot be a member of a class - and this restriction includes itself too. as i said this follows from zermelo's axioms. the axiom of separation (AS) states that

if A is a set then A' = {x∈A:P(x)} is a set as well.

less formally this means that

if B⊆A and A is a set then so is B

taking the contraposition of this implication, we get an equivalent statement:

(given that B is a subclass of A) if B is not a set then neither is A.

now, if the set of all sets existed, it would have an extension of the form

{x:x is a set}

applying the axiom of separation, we can single out some members and still get a subclass that is a subset. namely, take A' such that

A' = {x∈A:x∉x}

by AS, the subclass above is a set. but we've just written an extension of the russell class, which is clearly not a set. a contradiction. or you can look at it differently and note that since russell's class is included in A, by the contraposition of the AS, A is not a set. either way, A is not a valid object in the universe.

now, you might argue that russell's class doesn't really count due to its bizzare and somewhat artificial nature. amazingly enough we can show that the class is, in fact, perfectly legal. its existence follows from cantor's theorem:

if A is a set and P(A) is its power set, then |A| < |P(A)|

among other things, it implies that there does not exist a mapping from A into P(A) that is a bijection (ie, it's never surjective).

now, if A is the class of all sets, then P(A)⊆A. let us introduce an equivalence relation on A as follows: {(x,x):x∈A}. this diagonal relation (call it _R_) is clearly a bijection from A to itself. but look, we just showed that there is a map from A to a class that includes P(A) (namely A itself) that is one-to-one AND onto. and since |A| ≠ |P(A)|, the class

C = {x∈A:x∉_R_(x)}

becomes russell's paradoxical class.

Not to sound cocky, I'm fairly intelligent, but you Noisician make me feel like I know absolutely nothing.

[edit]Dang, a revolting devlopment from another thread...

Beer came first!

But...

AC/DC - Who Made Who



[[[smoke]]]

quote:

Originally posted by Noisician its existence follows from cantor's theorem: if A is a set and P(A) is its power set, then |A| < |P(A)| among other things, it implies that there does not exist a mapping from A into P(A) that is a bijection (ie, it's never surjective).

Cantor's theorem is nice. A similar application of his famous diagonal slash proves it too:

Suppose you have a set S, and a function f(x):S→P(S).

Define the set T = {x∈S | x∉f(x)}.

Now, since T ⊆ S, clearly T ∈ P(S). Now suppose f(x) is surjective. Then we can denote the pre-image of T as t∈S.

But then we have:
t∉f(t)=T→t∈T and
t∈T→t∉f(t)=T
A contradiction! Hence, T is not surjective, and f(x) cannot be a bijection from the a set to its power set.

I had never seen that use of Russell's paradox to invalidate the concept of a set of all sets, however. Maybe I'll go learn a bit of set theory

BTW, I've never seen the notation x∉_R_(x), where R is a binary relation. I understand it to mean ¬_R_(x,x) or something? Without understanding that, the final step in your explanation is hard to follow:

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Originally posted by Noisician and since |A| ≠ |P(A)|, the class C = {x∈A:x∉_R_(x)}

I take this to mean the following: since |A| < |P(A)|, and since _R_:A→P(A) and exhibits a bijection by construction, there exists x∈A that does not meet the trivial equivalence relation? I'd love it if you clarified here.


P.S. I think we're boring everyone.

fcuk, I barely passed that mathematics course and you bring up the memories yet again.. gah..

I would have to say the chicken came first.

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Originally posted by Flyboy217 BTW, I've never seen the notation x∉_R_(x), where R is a binary relation. I understand it to mean ¬_R_(x,x) or something? Without understanding that, the final step in your explanation is hard to follow:

in addition to being an equivalence relation, _R_ is also a function. hence, _R_(x) is the same as _R_x, whichever notation you are more familiar with. it's called the identity relation on a class. look at the following definition of a function - it should then become clear what i meant in my previous post:

� = {(x,�x):x∈dom �}

quote:

Originally posted by Flyboy217 I take this to mean the following: since |A| < |P(A)|, and since _R_:A→P(A) and exhibits a bijection by construction, there exists x∈A that does not meet the trivial equivalence relation? I'd love it if you clarified here.

you almost answered your own question with the "diagonal slash" you provided

let me restate it a little bit.

in order to show that |A| < |P(A)| we need show a) |A| ≤ |P(A)| and b) |A| ≠ |P(A)|

a) we define a map h from A into P(A) as follows: ha = {a} ∀a∈A. clearly, h is an injection from A to its power set.

b) we let m be a map from A to P(A). then for every x∈A, mx is a member of P(A) - that is, a subset of A. take

C = {x∈A:x∉mx}

then C is a subset of A - that is, a member of P(A). if m were surjective, there would be some c∈A for which mc = C. therefore c∈mc if and only if c∈C.

but from the definition of C above, c∈C if and only if c∉mc. a contradiction.

now, returning to my previous post, just take the identity relation on the class of all sets i described above as the m in the proof, and the C of the proof becomes russell's class of all sets that do not belong to themselves.