The search for absolute infinity

[matt grime]

Having read back a couple of posts, can I make the following observation without repeating someone else?

If you are going to have a universal set which has a metric (or topology) then the collection of all metric spaces will inherit that metric (topology) as a subset (subspace), and thus you come up against Russell's paradox straight away.

This is even something that physicists are finding in string theory and spin foam models.

Please point out if I'm way off topic, and I'll delete this straight away.

 

[phoenixthoth]

it's not off topic in my opinion. seems like the only way to equip U with a metric is with the trivial one. I'm not seeing how russell's paradox would be involved in the metric space aspect of it, though it is run up against in the main aspect of the existence of U. that all metric spaces would inherit the metric from U's metric leads me to suspect that the only metric definable on U is a trivial one in which
d(x,y)=1 if x!=y, else d(x,x)=0. and that would lead me to believe that the only topology on U that makes sense is for every set in U to be open.

russell's paradox is handled in the main treatment of U, as well as cantor's diagonal argument (somewhere near theorems about P(U) not being bigger than U). cantor's diagonal set boils down to russell's set in a certain situation.

incidental note: in quinne's (quine?) new foundations theory with a universal set, he somehow manages to avoid russell's paradox though without using three valued logic and with the axiom of choice being false for some reason. I'm guessing that's why not everyone has heard of it: no axiom of choice means no zorn's lemma and many things crumble in various fields (pun not intended). so far, I've been unable to find a free copy of his works online though a book in 1995 with his ideas is only $35. i would buy it to look for more theorems about U to prove.

 

[Russell E. Rierson]

The goal is to eliminate paradox, while maintaining an all inclusive principle of "comprehension"[semantics], yo, where an infinitely expanding chain of "sets[containment principles]" and concepts, such as "proper set", "ordinal" and "cardinal" are relativised to context, which would take care of paradox at all levels, except for the "top", which naturally does not exist, of course! So it becomes an infinite chain or composition of ever more inclusive situated sets[semantics] with an interesting informational - topological dynamic. So it comes full circle, and the poetic verses explaining Beingness and Nothingness become a unifying dialectic, and a new synthesis. It just needs to be put into a rigorous mathematical framework[syntax].

Barwise Situation Theory?:

http://www.cs.bilkent.edu.tr/~akman/jour-papers/sigart/node1.html

 

[Russell E. Rierson]

Interesting:

http://www.cs.bilkent.edu.tr/~akman/jour-papers/air/node13.html

Barwise defined the operation M (to model situations with sets) taking values in hypersets and satisfying (cf Note 12):


if b is not a situation or state of affairs, then , M(b) = b

if , rho = <R,a,i> then M(rho) = <R,b,i> (which is called a state model), where b is a function on the domain of "a" , satisfying ,
b(x) = M(a(x))

if s is a situation, then M(s) ={M(rho) : s |= rho}.

Using this operation, Barwise then proves some theorems, including the one which states that there is no largest situation (corresponding to the absence of a universal set in ZF).

 

[matt grime]

If you are going to have a universal set, and if the collection of *all* metric spaces is a subset of this set, then a metric on the universal set would imply a metric on this set of all metric spaces, and you have a set that contains itself. The way to sidestep it would be to say that the universal set contains only *a* set of metric spaces, not _the_ set of all metric spaces. Personally I adhere to the Grothendieck school and just ignore these issues.

As for Quine's Universal set theory, I can see how the assumption that Zorn's Lemma is false would be used. Without it, P(U) might be empty.

 

[phoenixthoth]

my gut tells me that deduction operators like |= wouldn't apply the same way in ternary logic.

Originally posted by matt grime
If you are going to have a universal set, and if the collection of *all* metric spaces is a subset of this set, then a metric on the universal set would imply a metric on this set of all metric spaces, and you have a set that contains itself. The way to sidestep it would be to say that the universal set contains only *a* set of metric spaces, not _the_ set of all metric spaces. Personally I adhere to the Grothendieck school and just ignore these issues.

As for Quine's Universal set theory, I can see how the assumption that Zorn's Lemma is false would be used. Without it, P(U) might be empty.

this doesn't seem to be a problem because every set can be equipped with a trivial metric. one way to say that the reason for this could be that this is the only way to metrize U in which case all sets can be equipped with a metric inherited from U, being the trivial metric that basically says each point "knows" if a second point in question is "me" or "not me."

by changing the foundation axiom, i have a statement that wouldn't apply if there were no universal set and when there is one, it is a member of itself.

as far as the powerset of U goes, P(U)=U. furthermore, i have a couple of other theorems in my paper which go something like this:
1. P(x)=U iff x=U.
2. if x is crisp then there is a map from x onto P(x) if and only if x=U.
3. if P(x)=x then either x is fuzzy or x=U.

the first statement implies that U will never be built from below by powerset operations, which is something already known. i believe stronger versions of 2 and 3 are lurking out there that would say:
2'. there is a map from x onto P(x) iff x=U
3'. P(x)=x iff x=U.
i believe those are out there because in my theory, the powerset ignores partial membership and is somewhat "forgetful" with respect to elements that are only partial members.

cantor's diagonal argument which would normally rule out 2 and 2' i believe i have dealt with by adapting the subsets axiom in a way that
a. accommodates ternary logic
b. in a way that extends the situation for binary logic so that
if all one wanted to do was regular set theory with the two new axioms then nothing should be different.

 

[Russell E. Rierson]

http://www.cs.bilkent.edu.tr/~akman/conf-papers/Tueb/node6.html

One can assert facts that a situation will support. For example, if s1 supports the fact that Bob is a young person, this can be defined in the current situation s as:
s: (|= s1 (young Bob)). Note that the syntax is similar to that of Lisp and the fact is in the form of a predicate. The supports relation, !=, is situated so that whether a situation supports a fact depends on where the query is made.

s1 supports relation A, defined as situation s

s: |= s1


U[]U[]U][]U[]U[]U...

"like points on a line" ...?

 

[phoenixthoth]

i was suggesting that the following looks like a line in which U is like a point on that line:
...U&isin;U&isin;U&isin;U&isin;U&isin;U...
yet it's kind of a hyperline in that those six dots mean a lot of stuff (transfinite membership strings and such).

 

[phoenixthoth]

U as a pseudo-pseudo metric space:
define d to be a function from UxU to U (not just R) such that if x and y are sets, then
d(x,y)=x+y.

then
1. d(x,y)=0=Ø iff x=y
2. Ø<=d(x,y) where <= can be taken to mean either "is a subset of" or "can be injected into"
3. d(x,y)=d(y,x)
4. d(x,y)=d(x,z)+d(z,y).

then I'm wondering about limits and open balls.
an open ball would be something like this where delta is a set:
B(x,delta)={y in U | x+y<=delta}.

lim_n->Ux_n=x means that for all nonempty e, there is a set N such that x_n+x<=e whenever n<=N. i haven't decided which of these <='s it would be best to have as subset and which as "can be injected into." i'll have to do some examples like [/sub]x_n=U for all n and [/sub]x_n={0,1,2,...,n} or =(0,1/n) or =[0,1/n) or =[0,1/n] modulo some adjustments because n can be any set. the adjustment might look something like this: [/sub]x_n={0,1,...,n} if n&isin;N and [/sub]x_n=N (the set of natural numbers), or some infinite set, otherwise.

anyways, this "metric" is kind of neat in a way. if you visualize two sets and consider + to be xor (symmetric difference) then the more the sets have in common, the closer to empty this "metric" is; ie the closer the two sets are. the less that have in common or the larger they are, the further away from empty (0) the "metric" is.

if cosine can be extended to hyperreal numbers, can it be extended to arbitary sets? perhaps an angle between sets, which would be a set like all other angles just not a real one, is possible. I'm guessing that the "angle" between sets will be super in multiplicity and not constant modulo some ideal. hmmm... maybe the angle should be some coset x+J where cos(J)=1=U and cos(x)=1. arccos can be defined in terms of logs which can potentially be defined as an inverse to the powerset operation. thought for food...

 

[Hurkyl]

The hyperreal numbers have an extremely special relationship to the reals that enable you to transfer real functions to hyperreal functions; I don't think you can do anything here. It feels weird wondering about the angle between two sets as well, instead of looking at dot products...


Have you considered having d(x, y) mean |x + y|? I.E. the cardinality of the set (x + y)?


Also, what about multisets? We can make a module of multiset-like things over the cardinals, and then we can make a dot product out of the metric, and given a dot product we can define the angle between multisets.


We might be able to do the same with ordinary sets, but the base field would be Z_2, which puts our metric living in the "wrong" domain.

 

[phoenixthoth]

Originally posted by Hurkyl
The hyperreal numbers have an extremely special relationship to the reals that enable you to transfer real functions to hyperreal functions; I don't think you can do anything here. It feels weird wondering about the angle between two sets as well, instead of looking at dot products...


Have you considered having d(x, y) mean |x + y|? I.E. the cardinality of the set (x + y)?


Also, what about multisets? We can make a module of multiset-like things over the cardinals, and then we can make a dot product out of the metric, and given a dot product we can define the angle between multisets.


We might be able to do the same with ordinary sets, but the base field would be Z_2, which puts our metric living in the "wrong" domain.

interesting...

for |x+y|, how would you prove the triangle inequality? i tried a little but got stuck.

how would you make a dot product out of the metric? seems like you might have this for the angle:
cos t = (a.b)/|a||b| but we'd have to divide cardinals wouldn't we?

all one has to do for a dot product is define |a| because then since |a|^2=a.a and a.b=((a+b).(a+b)-a.a-b.b)/2, we get a.b=(|a+b|^2-|a|^2-|b|^2)/2. if those are all infinite cardinal numbers, we get a.b=|a+b|-|a|-|b| though showing that that satisfies the definition of dot product is probably not possible. but it's kind of pseudo-dot like.

thanks for the suggestions.

 

[Hurkyl]

Well, it's clear that |x U y| <= |x| + |y|, right? And we have x + y is a subset of x U y, so we have |x + y| <= |x| + |y|


The cardinal arithmetic is messy; I don't know if anything can be done with it.

 

[phoenixthoth]

lim_n->Ux_n=x means that for all nonempty e, there is a set N such that x_n+x<=e whenever n<=N.

made a mistake. i should have said N<=n. but i'll use the letter y or n' because of the possible confusion with the set of natural numbers.

btw: x_n would be a function from U to U.

i'm wondering why e has to be nonempty. if it is "for all nonempty e" then i can get limits whose symmetric difference is either empty (implying equality ), a singleton, or a doubleton.

if i change that to "for all e" then i can get limits to be unique.

working on a cauchy completeness type of business now now that uniqueness is in order. also working on giving one darn example that's not trivial.

 

[phoenixthoth]

zorn's lemma

zorn's lemma:
let S be a nonempty partially ordered set (ie we are given a relation x<=y on S which is reflexive and transitive and such that x<=y and y<=x together imply that x=y). a subset T of S is a chain if either x<=y or y<=x for every pair of elements of x,y in T (ie every pair of elements of T are comparable). Then Zorn's lemma may be stated as follows: if every chain T of S has an upper bound in S (ie if there is an x&isin;S such taht t<=x for all t&isin;T) then S has at least one maximal element.

consider the relation <= given by x<=y iff there is a 1-1 map from x into y. then can zorn's lemma be strengthened so that instead of x<=y and y<=x implying x=y, it just imples that they are isomorphic (as sets)?

let's leave it the way it is and suppose that f is a map from A to B where A and B are subsets of U. let f[A] denote the image of A under f, ie f[A]={f(a)&isin;B: a&isin;A}.

suppose that f has the property that n<=m iff f(n) is a subset of f(m). i will abbreviate this by writing f(n)$f(m). also suppose that there is an M such that f(n)$M for all n&isin;A. one can assume that M is not U to get a stronger result.

what i want to show is that f "converges" to some limit in this universal limit sense. i believe i can show that lim_n->Uf(n)=L iff there is an n' such that for all n, if n>=n' then f(n)=L.

let S be f[A] u {M}. by the assumption on f, every element of S is comparable. i claim that any chain T in S has an upper bound in S. let T be a chain in S. then T's elements are comparable as all elements in S are comparable. every element in T has an upper bound in S: namely M. then S has at least one maximal element L.

either L=M or L&isin;f[A]. if L&isin;f[A], then i claim that f U-converges to L, ie that lim_n->Uf(n)=L. we know that f(n)$L for all n&isin;A as L is a maximal element of S. as L&isin;T, f(n')=L for some n'&isin;A. now suppose n>=n'. by assumption on f, f(n) contains f(n')=L; hence f(n)=L. by the lemma i haven't proved here, this is sufficient to prove the claim.

if M is not in f[A] the i claim f does not converge. i haven't worked out the details.

i'm trying to also show that U is a noetherian ring, ie a ring which satisfies the ascending chain condition. i want to use this result and take f(n) to be a n-th ideal of some kind which form an asencing chain. then M might be the union of the f(n)'s or something.

is it known that boolean rings are noetherian? some, all, none?

 

[phoenixthoth]

this is the current incarnation of it: http://www.alephnulldimension.net/matharticles/tuzfcver8-1-0.pdf (103kb/12 pages)

i'm contemplating publishing this somewhere but i'd appreciate feedback before i try to do so so i don't make a complete a** out of myself.
any thoughts?

 

[matt grime]

The proof of theorem 2B is wrong.

just because something is an element of the power set P(x) does not imply it is an element of x.

In fact, it might be that, if we can form the set of sets not equal to U that its power set is U (if it isn't U already). I didn't look at the ternary logic enough to state that for certain.


Also the proof that no proper subset of U is in bijection with U is wrong - as you are using ZFC with U, you have the axiom of infinity, which assures that there is an inductive set, hence U contains the sets used to define the infinite set, and thus there is s trivail bijection from U to U\{{}} that is U omitting the set containing the empty set. This is a constructive proof, so ternary logic doesn't enter into it.

 

[phoenixthoth]

...{a} is an element of P (x) implies that {a} is a subset of x. this implis that a is an element of x. since a was arbitrary, by the uniqueness of U, x=U.

"there is s trivail bijection from U to U\{{}} "
can you please prove that it is a bijection because i don't see that.

 

[matt grime]

First one: {a} in P(x) implies that a is a subset of x, it does not imply a is an element of x.

proof x is an element of P(x) but for an arbitrary set x is not in x.

there was no other constraint placed on x other than it be a set whose power set was the universal set.


second. if you have all the ZF axioms then you have a collection of sets labelled by the integers - the elements in the inductive set, send the set labelled by 1 to2, by 2 to 3 etc. and define the map to be the identity elsewhere. this is clearly a bijection onto a proper subset, and it works for any set containing an infinite number of elements, and it is constructive.

 

[phoenixthoth]

again, i wrote this:
...{a} is an element of P (x) implies that {a} is a subset of x. this implis that a is an element of x. since a was arbitrary, by the uniqueness of U, x=U.

not this:
...{a} is an element of P (x) implies that a is a subset of x. this implis that a is an element of x. since a was arbitrary, by the uniqueness of U, x=U.

oh, i get it now. in general Z in P(x) implies Z is a subset of x, no? let Z={a}.

anyways don't you expect this theorem to be true anyway because it implies that U can not be arrived at by power-setting a smaller set.

"second. if you have all the ZF axioms then you have a collection of sets labelled by the integers - the elements in the inductive set, send the set labelled by 1 to2, by 2 to 3 etc. and define the map to be the identity elsewhere. this is clearly a bijection onto a proper subset, and it works for any set containing an infinite number of elements, and it is constructive."

i'm not understanding the relevance. is that a map that is a bijection from U onto a PROPER subset of itself, U\{{}}? how can a map from a set to a proper subset of itself be injective? oh i see, in the case of potentially infinite sets of course! duh. but still, I'm not seeing an explicit example of a bijection between U and a proper subset of U which WOULD violate something in my paper. intuitively, you're stirring around the elements in U not mapping U bijectively to a proper subset of U. I'm just not understanding you but please please be patient with me.

 

[matt grime]

Erm, in what way did you not understand the counter example to the 'proof' you've got? (the unversal thing isn't at issue here, just the assertion that as {a} is in P(X), that a must be
*an* element of X. a is a collection of elements of X is all that you can deduce. X is an element of P(X) yet X is not in general an element of X!


Second. You what? By construction the map is injective, find distinct x and y with f(x)=f(y) for f the function defined in my last post. It's elementary to show that it is injective, unless you aer going to argue that I cannot split the universal set into those elements in the inductive set and those not.


A map from a set to a proper subset can easily be injective if a set is infinite as you yourself allude to in the paper round about that theorem on there being no bijection from U to a proper subset of itself. Remove the axiom of infinity and this example goes away.

 

[phoenixthoth]

i edited my post but it's not of real consequence now.

Originally posted by matt grime
Erm, in what way did you not understand the counter example to the 'proof' you've got? (the unversal thing isn't at issue here, just the assertion that as {a} is in P(X), that a must be
*an* element of X. a is a collection of elements of X is all that you can deduce. X is an element of P(X) yet X is not in general an element of X!

ok. let z be any set and consider P(x). suppose z is in P(x). is z a subset of x or not? suppose it is. then that means all elements of z are in x. ok? now take z={a}. since the assumption is that P(x) is the universal set, {a} is in P(x). then all elements of {a} are in x. that means a is in x.

Second. You what? By construction the map is injective, find distinct x and y with f(x)=f(y) for f the function defined in my last post. It's elementary to show that it is injective, unless you aer going to argue that I cannot split the universal set into those elements in the inductive set and those not.

let me get this straight. is this an equivalent example:
U=N union (U\N).
let f be a self mapping of U such that
f(x)=x+1 for x in N and
f(x)=x for x in U\N.
is that your example?

thanks for helping me correct my paper, btw.

 

[matt grime]

Originally posted by phoenixthoth
i edited my post but it's not of real consequence now.


ok. let z be any set and consider P(x). suppose z is in P(x). is z a subset of x or not? suppose it is. then that means all elements of z are in x. ok? now take z={a}. since the assumption is that P(x) is the universal set, {a} is in P(x). then all elements of {a} are in x. that means a is in x.

No, the assumption about the unversality of P(X) does not come into it. Firstly, {a} is a set with one element, a, that a is also a set is misleading. You cannot conclude that all the elements of {a} are in x. Secondly, the last line exactly states the objection that a is only a subset of x. You are confusing 'is a subset of a set' with 'is an element of a set of sets'

let me get this straight. is this an equivalent example:
U=N union (U\N).
let f be a self mapping of U such that
f(x)=x+1 for x in N and
f(x)=x for x in U\N.
is that your example?

thanks for helping me correct my paper, btw.

Not quite, I want a collection of sets labelled by N, not the set N itself. The axiom of infinity means that such must exist in any model we're looking at. The sets are:

0 - {} the empty set
1 -{{}} the set containing the empty set
2 -{{},{{}}} the set containing the two previoius sets.


I can just shift the labels by one here and leave all other sets unchanged.

 

[matt grime]

Acutally ignore the universal set power set unique thing for now, maybe some light has just come on in my head.

As written your proof could do with explanation, well, as written the 'proof' is wrong or at least the assertion needs more explaining, but the result might hold.

I believe you want to consider the set which contains the set which contains a, for then {a} is a subset of X, but it contains one element, then a is in X - too few braces used

 

[phoenixthoth]

"Not quite, I want a collection of sets labelled by N, not the set N itself. The axiom of infinity means that such must exist in any model we're looking at. The sets are:

0 - {} the empty set
1 -{{}} the set containing the empty set
2 -{{},{{}}} the set containing the two previoius sets.


I can just shift the labels by one here and leave all other sets unchanged."

i'm only detecting the essence of what you're saying but I'm not getting it just quite yet. the claim, your claim, is that there is a bijection between U and a proper subset of U. are we discussing this corollary: if x is a proper subset of U, then there is no 1-1 map from U to x? I've lost track because I'm having a brain fart and you shot off two points before i could handle the first one. overload! so, if we're discussing that corollary, then your example should indicate (hopefully as explicitly as i need it to be if possible) a set x that isn't U such that there is a 1-1 map from U to x. i think your claim is that there is a 1-1 map from U to U\{{}}. i do want to keep the axiom of infinity (especially since i think it might be a consequence of the universal set axiom and so i must keep it), so i want to exactly pinpoint my error. well, i admire you if this example is 'trivial'.

 

[matt grime]

First, the P(X) thing can be corrected, as hopefully you saw above:given any set Z, {{Z}} is in P(X)

so the set with one element {Z} is a subset of X, so Z is an element of X.



The second.

Yes, it is your corollary that there is no 1-1 map form U to a proper subset of U, and the statement that for all sets A, U\{A} is not in bijection with U. Well, that isn't true if you have the axiom of infinity.

 

[phoenixthoth]

doesn't {{Z}} is in P(X) imply that {{Z}} is a subset of x?

 

[matt grime]

Originally posted by phoenixthoth
doesn't {{Z}} is in P(X) imply that {{Z}} is a subset of x?

Yes. It implies that {{Z}} is a set of some things in X, right? ie that {Z} is some set of elements of X, but {Z} has only one element, Z, so Z must be an element of X. Careful with your bracketing.

 

[phoenixthoth]

"Yes."

so if {{Z}} is a subset of x then the following conditional holds for all sets y:
if y in {{Z}} then y in x. ok?

suppose y in {{Z}}.

first of all, that means y in x.

second of all, that means that y={Z}. this works the other way: {Z} is in {{Z}}.

hence, {Z} in x.

now go back to {a} is a subset of x (which follows from {a} is in P(x)):
for all sets y, if y in {a} then y in x. ok?

suppose y is in {a}.
1. y=a
2. y in x.
3. therefore, a in x.

it's ok as it is.

 

[matt grime]

Maybe it's me that's got his braces wrong, it's a headache, but i agree the result is true. I don't dispute the result, and I increasingly think I agree with your original proof.

I stand by the second issue though, about proper subsets

 

[phoenixthoth]

i will heed your advice about being careful though.

i'd still like you to be more specific and concrete with me on your counterexample because that would seriously damage the paper. i mean to be as detailed as possible because I'm not an expert in set theory so i can't understand your sketch.

 

[matt grime]

By the ZF axioms there is the empty set {}, then there is the set containing it { {} }, and then the set containing those { {} , {{}} }, and so on, each of these can be labelled by an element of N correspeonding to the cardinality. The axiom of infinity states, that when I say 'and so on' that actually there is an infinite number of sets created inductively (this apparently does not follow from all the other axioms) with the labels from all the natural numbers. If you have this, your universal set cannot be finite, and must contain these sets.


Since there is a bijection from N to a proper subset of itself (n to n+1) then there is a bijection from those sets to a proper subset of the sets, and defining it to be the identity for all other sets gives a contradiction to your corollary.

 

[phoenixthoth]

thank you.

now another question.
that corollary was meant to be the contrapositive to the statement: if f is a 1-1 function from U to x, then U = x. what's the contrapositive of that? because if the contrapositive is wrong that means the theorem it draws its energy from is wrong.

 

[phoenixthoth]

what are "those sets " you mention in paragraph 2? they're indexed by N right? but what are they? are they N, P(N), P(P(N)), etc., which can be indexed by 0, 1, 2, ...? just for reference, here is the axiom of infinity:
$$\exists x\left( \emptyset \in x\wedge \forall y\in x\left( y\cup \left\{ y\right\} \in x\right) \right)$$

 

[matt grime]

I said above

0 ---{} the empty set
1---{ {} } the set containing the empty set
2 ---{ {} , { {} }} the set containg the previous two sets

3 is the set continaing the previous 3 sets, and so on n is the set containing all the previous n-1 sets constructed. This is often called omega, a quick google for axiom of infinity wil provide you with some good links.

 

[phoenixthoth]

omega is the ordinal number for the set you're describing, which is N.

0 is defined to equal Ø.
1 is defined to equal {Ø}.
2 is defined to equal {0,1}
3 is defined to equal {0,1,2}.
N is defined to equal {0,1,2,...} which exists by the infinity axiom.
do a search on that yourself. see enderton's "elements of set theory," et al.

so your map really is this:
f(x)=x+1 for x in N
f(x)=x for x in U\N.
that's what i thought.