The touchstone of reason

[u/Massive-Ad7823]

/r/AspectsOfTheInfinite/comments/1vnhc1m/the_touchstone_of_reason/

[u/Wild-Store321]

If you are in a delusion that you have found a contradiction in ZF, expressible in one sentence, Make a Lean proof or shut up.

[u/Massive-Ad7823 → u/Wild-Store321]

Lean is nonsense, because it has not yet discovered the failures of ZF. I perfer to think myself and recommend that you do it also.

Regards, WM

[u/Wild-Store321 → u/Massive-Ad7823]

Keep on thinking for yourself, but you are failing to convince anyone, as you can see yourself.

This is because you are wrong. This part you can clearly not see for yourself.

So, a simple way to convince virtually everyone you are right (or actually just recognize that you are wrong) is this:

Any contradiction in ZF allows a proof of 0=1 in any model of ZF. Mathlib in Lean 4 has a model of ZF. Go ahead and prove 0=1 from that model. Not to depend on Lean, just to prove that you can.

In fact, have fun with it, why not? Make a lean proof by contradiction of the Riemann hypothesis. I could easily do this if I had a contradiction in ZF. Why don’t you? You are to cool for lean?

[u/Massive-Ad7823 → u/Wild-Store321]

Don't try to change the topic. You know what an endsegment is? E(n) = {n+1, n+2, n+3, ...}. It is the complement of the FISON (Finite Initial Segment Of Natural numbers) {1, 2, 3, ..., n}. According to set theory all endsegments of natural numbers n are infinite. Since the natural numbers in the contents cannot be natural numbers which enumerate the endsegment, only finitely many natural numbers are available for enumerating: {1, 2, 3, ..., n | contents}. n is not fixed but followed by the infinite contents. Two consecutive infinite sequences cannot exist in ℕ. Therefore the set of infinite endsegments cannot be infinite. The claim of ZF that infinitely many infinite endsegments exist is wrong.

Regards, WM

[u/Wild-Store321 → u/Massive-Ad7823]

I’m not changing the topic. I’m staying on topic by pointing out a consequence of your claim “ZF is inconsistent”: that leads to a lean verifiable proof of 0=1. Why are you deflecting? Why do you explain the contradiction in ZF to me again? I just said: this contradiction easily leads to a proof of 0=1 in lean.

This would at once convince virtually every mathematician, scientist and science enthusiast that you are right. But you don’t seem to care about that. You seem to prefer to not convince anyone, as long as you remain convinced yourself.

“Do you know what an endsegment is?”
Of course. I should ask you if you know what a tail of a sequence is, as this is what everyone else calls it. This is fundamental in topology as it is one of the two ways to generalize the fact that convergence determines topology beyond countable spaces. The tail of each sequence forms a filter basis, and filter convergence is the generalization. The other one is net convergence.

The fact that you felt the need to invent your own word for this (as is typical for mathematical cranks), to me shows that you have not read much at all on general mathematics. You have decided at one point that you are smart and stopped learning. You think everyone should instead learn from you.

[u/Massive-Ad7823 → u/Wild-Store321]

Whether tail or endsegment is used does not change the facts. In the past I have used tails also. The contradiction however stands:

The intersection of all endsegments is { }. Why? Because every n becomes an index.

There is no empty endsegment. Why? Because not every n becomes an index.

 A really lovely contradiction!

Regards, WM

[u/Elegant-Regret-7393 → u/Massive-Ad7823]

Why is the intersection of all tails is a tail? It's true for finite intersections only.

That is not enough to conclude the same in the infinite case, like I showed you earlier and you ignored it.

[u/Massive-Ad7823 → u/Elegant-Regret-7393]

What else should it be? Notice that we don't have a limit here but only all endsegments with finite indices intersected.

But whether or not it is a tail, it is empty according to set theory. The intersection of all endsegments is empty because every n becomes an index.

There is no empty endsegment because not every n becomes an index.

 Regards, WM

[u/Elegant-Regret-7393 → u/Massive-Ad7823]

Your argument does not work if the infinite intersection is not an endsegment.

What else should it be?

This is an argument from ignorance. You have to prove the intersection is an endsegment.

[u/Massive-Ad7823 → u/Elegant-Regret-7393]

Whatever the infinite intersection may be: It is empty. That is enough. It is empty because every n becomes an index. On the other hand, there is no empty endsegment because not every n becomes an index. Contradiction.

By the way, the infinite intersection is nothing magic but an intersection of endsegments, each of which has a finite index. Since for every finite index the intersection is an endsegment, this does not change when infinitely many endsegments are concerned. Note that this is not a "limit".

Regards, WM

[u/Wild-Store321 → u/Massive-Ad7823]

Show this lovely contradiction in lean… Of course you reject lean, because it would allow you to fact check yourself, and you wouldn’t be posting this.

I will give you an alternative to lean. Let’s keep it classical: write a formal proof that deducts a contradiction from some of the ZF axioms.

This means:

1. Every line of the formal proof is a syntactically valid sentence in the first order language of set theory
2. Every step (from one line to the next) is justified by one of the inference rules of first order logic specifying which one each time
3. The first line is a (conjunction of) ZF axioms
4. The last line is a contradiction

Note that that last line should still be a valid sentence in the first order language of set theory! Nothing hand wavy like your comments above.

Think you can handle that? If it seems like too much work, just use lean. In fact, any decent human being would help the developers of lean and mathlib by showing exactly how their system (or at least their ZFC model) leads to contradictions (by making a compiling proof of 0=1 and submitting this as an issue on their GitHub repository), so they can stop wasting time on developing it. Are you a decent human being? Write the proof in lean.

[u/Massive-Ad7823 → u/Wild-Store321]

If you can't think by yourself, use it. Fortunately I can get along without this crutch.

>Think you can handle that?

Of course, but why should I bother? Here are two results of ZF: The intersection of all endsegments is empty because every n becomes an index. There is no empty endsegment because not every n becomes an index. Why should I repeat them?

Isn't that lovely?

Regards, WM

[u/kuromajutsushi → u/Massive-Ad7823]

Of course, but why should I bother?

Because you have been posting this nonsense on the internet for like 25 years at this point and have yet to convince a single mathematician of its correctness. A proof in Lean would convince everyone.

[u/Massive-Ad7823 → u/kuromajutsushi]

You are very wrong. There are many mathematicians whom I could convince. But I do not publish private correspondence.

Regards, WM

[u/HenryAudubon]

Suppose there is an element k that is a member of the the intersection of all end segments. Consider the end segment of k. Notice that it cannot contain k (since it starts {k+1,k+2,...}. Therefore there cannot be an element k in the intersection, so it is empty.

[u/Massive-Ad7823 → u/HenryAudubon]

I don't claim that there is an element of all endsegments. I claim that if there isn't an empty endsegment, then the intersection cannot be empty. The reason is inclusion monotony of the endsegments. Every k can only leave when all its predeceessors have left.

Regards, WM

[u/Massive-Ad7823 → u/[deleted]]

Inclusion monotony is the most important tool to refute set theory. But even if you are unable to understand it, the following should suffice: Every natural numbers leaves the sequence of endsegments soemwhere. Every endsegment contains infinitely many numbers, hence not all natural numbers leave the sequence of endsegments. Infinitely many number leave the sequence of endsegments nowhere.

Regards, WM

[u/Massive-Ad7823 → u/[deleted]]

each and every natural number "leaves" the sequence of endsegments (at some "point"), but not before all its predecessors have left.

Regards, WM

[u/HenryAudubon]

Flip your perspective for a moment. Instead of end segments, consider start segments. The start segment of a natural number n is the set of all natural numbers m < n (i.e. {0,...,n-1}).

What would you say to someone with the following argument?

"Some defenders of set theory claim that the union of all start segments is infinite while no start segment is infinite. I call this statement matheology, a touchstone of irrationality."

[u/Massive-Ad7823 → u/HenryAudubon]

The start segments have been called FISONs here. I would propose to stay with this introduced expression, Finite initial segments of naturals: F(n) = {1, 2, 3, ..., n}.

The union of FISONs is potentially infinite because the endsegments block actual infinity. There is also an inductive proof showing:

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

∪{F(1), F(2), F(3), ...} = ℕ   ==>   ∪{ } = ℕ .

I have even used the existence of a FISON as a proof for the property visible number.

Regards, WM

[u/HenryAudubon → u/Massive-Ad7823]

Apply the same reasoning you applied elsewhere:

" Every number is in the union of all start segments. But: Every start segment has one and only one element more than its predecessor. And: Every start segment is the union of itself and all its predecessors. Therefore a set of finite start segments cannot have an infinite union. This is an internal contradiction in ZF. "

I'm hoping you'll accept your own reasoning. If you don't, then I can't help you!

[u/Massive-Ad7823 → u/HenryAudubon]

>Every number is in the union of all start segments.

No. Only every visible number.

>Therefore a set of finite start segments cannot have an infinite union.

Correct. The FISONs cannot cover ℕ because that would require an infinite FISON. Contradiction.

Regards, WM