1. Take a set {P}. If it's impossible to make this set a member of another set, then {P} always. Any attempt {{P}} will result in {P}

2. P = {P} where P is the set that contains itself. Let's try to make {P} a member of another set like this, {{P}}. But {{P}} = {P}. In other words, {P} always

So,

A set that cannot be made a member of a another set (see 1 P always) is the same as a set that can be made a member of itself (see 2 P always).

Contradiction.

Hence, Impossible/false that P = {P}.

am I to think that putting, say, a list of items e.g. 1, w, # inside curly braces like so, {1, w, #} amounts to doing nothing? — TheMadFool

You are to think that "doing something" is not a set theoretic predicate.

And you skipped my counterexamples to your incorrect reasoning.

this means that we agree to disagree on this. — Philosopher19

Circular. I asked what is the operative import of "agree to disagree". Your answer is that we "agree to disagree".

(1) Set theory is incomplete, therefore set theory is consistent.
— TonesInDeepFreeze

If that's what you want to believe, then believe. — Philosopher19

It's not just what I believe, it's easily proven, basically from the definitions of 'incomplete' and 'consistent'.

Rejection of the set of all sets is blatantly contradictory. — Philosopher19

You keep repeating that, but without showing an actual contradiction. That is dogmatism.

You are like child in your reasoning and manner of discussion. — Philosopher19

You've shown nothing childish. Especially nothing childish in informing you of the exact mathematical formulations that rebut your ignorant dogmatism.

I shouldn't have to spoon feed you — Philosopher19

Instead, you shove nonsense down the throat.

And I shouldn't have to spoon feed you clear and correct information about the subject on which you ignorantly espouse, but I do.

By definition, the set of all sets encompasses all sets. — Philosopher19

By the axioms, there is no set x such that every set y is a member of x.

That's not childishness; it's axiomatic mathematics.

Take a set {P}. If it's impossible to make this set a member of another set — TheMadFool

{P} is a member of other sets.

Meanwhile, as I already mentioned, without regularity, it is not inconsistent that

P = {P} = {{P}} = ... for finitely many iterations as you like.

A statement of set theory is inconsistent it it implies both a statement S in the language of set theory and the statement ~S.

x = {x} is inconsistent in set theory.

It is not the case that x = {x} is inconsistent in set theory without the axiom of regularity.

"something done" is not in the language of set theory, so it's not part of any statement S in the language of set theory. "always" is not in the language of set theory, so it's not part of any statement S in the language of set theory.

/

One thing ridiculous about your arguments is that they simply overlook that you get everything you want anyway just by recognizing that set theory has the axiom of regularity. That is, set theory agrees with you that no set is a member of itself. But it agrees with you via an axiom. But, as I have been making this point that you refuse to understand, if we drop that axiom, then it is not inconsistent for a set to be a member of itself. Of course, a set being a member of itself is intuitively incorrect to most people, which is fine. But inconsistency is not determined by what is intuitively incorrect, but rather by the definition: a set of formulas is inconsistent if and only if the set proves a contradiction, where a contradiction is the conjunction of some statement (in the language) and its negation.

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A good mirror for us all — tim wood

I'm all for self-scrutiny, but not so overzealous that I chase demons that there is no reason to think exist. I have plenty of faults, but being a crank is not one of them.

I get it that the set of all sets that are not members of themselves yields a paradox resolved by ruling that such a "set" is not a set under the rules. — tim wood

~ExAy(yex <-> ~yey) is proven from first order logic alone; we don't need any set theory axiom for that.

metaphorical — tim wood

Everything east is in the class, not set, of all sets. However, in Z set theories, that class itself is not an object; rather we refer to that class as an informal locution from "outside" the theory. But in class theories (such as NBG) the class of all sets is an object, but then there is no class of all classes that is an object in the theory.

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Every several thing east is a set in itself, but the collection of them, in the east, is not a set? — tim wood

Right.

{P} is a member of other sets. — TonesInDeepFreeze

Assume that it's impossible to make P a member of another set i.e. {{P}} is not possible or, more relevantly, always {P}.

Now if sets csn contain themselves, {P} = {{P}} = {{{P}}} =...

In other words, always P.

That means a set {P} that can't be contained in another set (always {P} never {{P}}) is the same as the set {P} that contains itself (even if {{P}} always {P}).

You see the point don't you? A set {P} that contains itself is the set that can't be a member of another set!

A set {P} that contains itself is the set that can't be a member of another set! — TheMadFool

That is so blazingly incorrect that it scorches the core of this planet.

That is so blazingly incorrect that it scorches the core of this planet. — TonesInDeepFreeze

:ok: Thank you for your time.

Thank you for your time. — TheMadFool

So you figured why this is not the case?:

without regularity
Ax({x}e{x} -> Ay(~y={x} -> ~{x}ey))

So you figured why this is not the case?: — TonesInDeepFreeze

No.

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If there is a set of all sets, then that set has a subset that is the set of all sets that are not members of themselves, which implies a contradiction.

No. — TheMadFool

You thought about it for at least half a minute?

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Imagine a high security prison, like a set {...}. You're the prison warden and under your careful watch, there have been no untoward incidents - no prison breaks, fairly good behavior from the inmates, no riots, etc.

One day, you're at your desk and a prison guard comes up to you and announces the arrival of a new prisoner, his name is K. K is a very unique prisoner. What's unique about fae? Well, fae has the following relationship with prisons, K ={K}. In other words, K inside the prison {K} is equal to (is the same as) K outside the prison.

You don't think that's too much of a problem and imprison K like so, {K} and feel quite content with the arrangement. In what seemed almost instantaneous, a guard informs you that K is no longer in prison because K = {K}.

You're angry and surprised but you give the matter some thought and realize K = {K}. You decide that one "solution" is to build another prison outside your prison and if so {K} will become {{K}} and K will become {k}. K is now in prison or so you think.

{{K}} = {K} but {K} = K. The prison guards are mortified as K is again outside the prison (it couldn't be contained in a set).

You then decide to build a third prison like so {{{K}}} and feel confident that you've finally managed to solve the problem...once and for all.

Unfortunately for you, {{{K}}} = {{K}} = {K} = K. Again, K is no longer in prison.

You hold a meeting with your colleagues and after many, many hours of brainstorming you soon realize the gravity of the situation K = {...{...{...{K}...}...}...} i.e. even if you build an infinite number of prisons, K can't be imprisoned.

You thought about it for at least half a minute? — TonesInDeepFreeze

Well, sorry. I have issues. I hope you'll cut me some slack and let my impudence slide. G'day. You've been very helpful.

But on our construction, the demon already sniffed that out and left it in the west as not a set. — tim wood

Then the demon is not allowing the subset operation. So the collection would not be one recognizable as serving an ordinary set theoretic role. But you do continue to say we'll disallow certain subsets:

But for the rest, there can be a set of all the other sets? — tim wood

Nope, if have taking of subsets, but then stipulate that we are not allowing in particular a set of all sets that are not members of themselves, then we could still derive a contradiction from "the set of all sets except the set of all sets that are not members of themselves".

Moreover, this demon would not be making his decisiond by algorithmic determination; his sniffing is purely magical, unlike set theory.

K inside the prison {K} is equal to (is the same as) K outside the prison. — TheMadFool

Set theory does not have a predicate "inside braces".

I can't imagine you'd have issues so severe that you couldn't see that self-membership does not imply impossibility of being in another set.

https://thephilosophyforum.com/discussion/comment/546862

I had a typo of omission there. I fixed it now.

And I said that even if the demon throws out the set of all sets that are not members of themselves, then there would still be a contradiction, if we allow subsets. I should add that what happens is that we get an infinite process of forming subsets and then the demon throwing them out ad infinitum. But then that is not analogous to set theory, even if we grant a premise of infinite process.

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his naive inchoate practices approximate those of logicians c. 1920. — tim wood

Which logicians and what formulations do you have in mind?

Nope, if have taking of subsets, but then stipulate that we are not allowing in particular a set of all sets, then we could still derive a contradiction.
— TonesInDeepFreeze
And this I do not see. — tim wood

I later corrected my typo of omission there.

It should be:

If we have taking of subsets, but then stipulate that we are not allowing in particular a set of all sets that are not members of themselves, then we could still derive a contradiction.

Do you want the proof?

It seems as I read it that you derive a contradiction from the idea of subsets in themselves. Am I misreading? — tim wood

Yes, you are. The axiom schema of separation is not inconsistent itself. It is inconsistent with the claim that there is a set of all sets.

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Keeping in mind that our (eastern) set has been scrubbed and disinfected of self-contradictory sets? — tim wood

A sentence (such as set existence assertion) is found to be contradictory with other sentences by being found to be a self-contradiction (logically false) or by being found to contradict previously proved sentences. So what the demons keeps or throws out, will be based on what he's already kept.

But set theory can't do that and be recursively axiomatized, because there is no algorithm to determine whether a given sentence contradicts a previous set of statements. That's why I say the demon is magical, because he's able to make immediate determinations that are not calculable even in principle.

It is true that there are maximally consistent sets of sentences. But membership in such sets is not algorithmically decidable.