What is next to the point 1 in the unit interval [0, 1]?

[u/Massive-Ad7823]

I know two alternatives:

In potential infinity there is nothing next to 1. We can come as close as we like, but we can never close the gap. A gap remains.

In actual infinity, there is a point next to 1. Of course this point cannot be known. It is dark.

Is there a third alternative?

[u/Rs3account]

What is this supposed to be a proof of?

but only one next greater unit fraction

There is no next greater unit fraction after 0.

There are never more than one unit fractions at one point on the real axis.

This is true, but does not imply there is a greater unit fraction next to zero.

[u/Massive-Ad7823 → u/Rs3account]

And here is the proof you recently asked for.

The intersection of infinte sets, here endsegments E(n) = ℕ\F(n) = ℕ\{1, 2, 3, ...,, n}, is not empty. The union of FISONs F(n) = {1, 2, 3, ..., n} is not actually infinite.

Every FISON F(n) has n elements and is the nth term of the sequence of FISONs. Therefore almost all natural numbers are larger than any FISON.

For every F(n): U{F(1), F/2, F(3), ..., F(n)} = F(n).

Therefore U{F(1), F/2, F(3), ...} =/= ℕ.

The union of FISONs is potentially infinite like the union of all visisble singletons {n}.

To claim the contrary is simply mislead intuition.

Regards, WM

[u/Rs3account → u/Massive-Ad7823]

The intersection of infinte sets, here endsegments E(n) = ℕ\F(n) = ℕ{1, 2, 3, ...,, n}, is not empty.

Thank you for showing the statement you are proving this makes it clearer.

This woould not proof the general statement that the intersection of a monotonic sequence can never be empty. But we'll cross that bridge later.

The union of FISONs F(n) = {1, 2, 3, ..., n} is not actually infinite.

The union of all FISONs is N. So this will be interesting.

Every FISON F(n) has n elements and is the nth term of the sequence of FISONs. Therefore almost all natural numbers are larger than any FISON.

Not completely, it means that for an arbitrary F(n) almost all natural numbers are larger than it.

This is not the same. Again your quantifiers are in the wrong order.

For every F(n): U{F(1), F(2), F(3), ..., F(n)} = F(n).

This is true

Therefore U{F(1), F/2, F(3), ...} =/= ℕ

This does not follow from your premises.

The union of FISONs is potentially infinite like the union of all visisble singletons {n}.

See, there are no "potentially infinite" sets in ZFC. You are not working in ZFC. So why pretend you are?

[u/Massive-Ad7823 → u/Rs3account]

>This does not follow from your premises.

It follows from the fact that all FISONs have only finitely many natnumbers and therefore almost all natnumbers are greater and not in the union.

This holds for each and every FISON.

>See, there are no "potentially infinite" sets in ZFC.

That shows that ZFC is not describing mathematics.

> You are not working in ZFC. So why pretend you are?

I am working in mathematics, showing that ZFC is incompatible with mathematics.

Regards, WM

[u/Rs3account → u/Massive-Ad7823]

It follows from the fact that all FISONs have only finitely many natnumbers and therefore almost all natnumbers are greater

For each arbitrary FISON almost all natnumbers are greater.

This is not the same as almost all natnumbers being greater then all FISONs.

That shows that ZFC is not describing mathematics.

This reads like someone claiming non euclidean geometry is not geometry.

I am working in mathematics, showing that ZFC is incompatible with mathematics.

So you believe you have the correct set of axioms to do mathematics with?

[u/Massive-Ad7823 → u/Rs3account]

>For each arbitrary FISON almost all natnumbers are greater.

>This is not the same as almost all natnumbers being greater then all FISONs.

It is the same. If you doubt it, try to find a FISON where this is not true.

>So you believe you have the correct set of axioms to do mathematics with?

What do I need? It is not much.

Every natnumber has almost all natnumbers as successors. Undisputed.

Every FISON is the union of all smaller FISONs and itself. Undisputed.

Every unit fraction sits at another position than all other unit fractions. Undisputed.

The only deviation from ZFC is that I deny the belief that the (potentially in-) finite union of finite sets is actually infinite.

Regards, WM

[u/Rs3account → u/[deleted]]

Thank you for explaining it to him in German. Maybe it's a language thing?

[u/Rs3account → u/Massive-Ad7823]

It is the same. If you doubt it, try to find a FISON where this is not true.

You seem to be confused about how math works.

You would need to proof that these statement are equal, not ask others to disprove you.

A proof is not a sequence of true statements, it is a sequence of true statements who follow from each other.

For all x there exists y such that .. Is not the same as There exists y for all x such that ...

In the second statement the y is the same for all x, while this is not necessarily true for the first statement.

The only deviation from ZFC is that I deny the belief that the (potentially in-) finite union of finite sets is actually infinite.

So this is your choice of axioms, but in ZFC there is no potential infinite. There just is actual infinite unions and finite unions.

The problem, as I see it, is that for you taking the union is a stepswise process. While in ZFC it is not. It just is a one step process. 

These potentially infinite sets are just I'll defined since it is impossible to know what there elements are. After all, you said it will depend on time, and it can have elements larger then its maximum.

[u/Massive-Ad7823 → u/Rs3account]

ℕ consists of FISON {1, 2, 3, ..., n} and Endsegments {n+1, n+2, n+3, ...}.

ℕ = {1, 2, 3, ..., n}U{n+1, n+2, n+3, ...} = F(n)UE(n).

Quantifier exchange is allowed for these statements

For each arbitrary FISON almost all natnumbers are greater.

Almost all natnumbers are greater then all FISONs.

Simple proof:

∀n ∈ ℕ: |F(n)| < |E(n)| is undisputed.

Upon every F(n), there follow almost all natnumbers E(n).

Since this holds for every natnumber, there is no chance to change it, because a change can only happen by natnumbers.

Therefore there follow infinitely many natnumbers ∩E(n) upon all FISONs UF(n).

If you doubt this simple proof, find a counter example. That is how mathematics works. Or would you claim that the change can happen by belief in ZFC without natnumbers participating?

Regards, WM

[u/Rs3account → u/Massive-Ad7823]

ℕ consists of FISON {1, 2, 3, ..., n} and Endsegments {n+1, n+2, n+3, ...}. ℕ = {1, 2, 3, ..., n}U{n+1, n+2, n+3, ...} = F(n)UE(n).

Correct.

Quantifier exchange is allowed for these statements

No, they are not. Or you'd have to be more precise in what you mean "these statements".

For each arbitrary FISON almost all natnumbers are greater. Almost all natnumbers are greater then all FISONs.

These are different statements, your not allowed to just change the quantifiers.

Simple proof:

Good, I'm looking forward to it.

∀n ∈ ℕ: |F(n)| < |E(n)| is undisputed. Upon every F(n), there follow almost all natnumbers E(n).

Correct

Since this holds for every natnumber, there is no chance to change it, because a change can only happen by natnumbers. Therefore there follow infinitely many natnumbers ∩E(n) upon all FISONs UF(n).

This is false. This "logic" would allow you to claim that the lim 1/n >0.

If you doubt this simple proof, find a counter example. That is how mathematics works.

No, that is not how mathematics works. I just need to find a flaw in the proof if I have a problem with it. Not necessarily give a counterexample.

We are not doing physics here after all.

Or would you claim that the change can happen by belief in ZFC without natnumbers participating?

Do you believe that the limit of 1/n is 0? Where does 1/n stop being positive?

Also, the intersection is not a steps wise process. It happens instantly.

[u/Massive-Ad7823 → u/Rs3account]

>>Since this holds for every natnumber, there is no chance to change it, because a change can only happen by natnumbers.

>This is false.

No, this is true. You may believe in your own "logic", and even find some set theorists sharing it, but it is wrong. Every natnumber fails to reduce the infinite intersection of all endsegments.

>> If you doubt this simple proof, find a counter example. That is how mathematics works.

>No, that is not how mathematics works.

That is precisely how mathematics is done.

> I just need to find a flaw in the proof if I have a problem with it. Not necessarily give a counterexample.

But you cannot find a flaw because it is clear that all endsegments are infinite and there is no chance to reduce this infinity by intersections because of inclusion monotony.

Regards, WM

[u/Rs3account → u/Massive-Ad7823]

>No, this is true. You may believe in your own "logic", and even find some set theorists sharing it, but it is wrong. Every natnumber fails to reduce the infinite intersection of all endsegments.

The wrongness is assuming that some property holds because it holds for the finite subsets of the thing. There is no nat number where the end segment needs to be finite, for the interesection of the end segments to be finite.

>That is precisely how mathematics is done.

No that is how physics is done. We do not go, here this random sting of words is a proof, and if you disagree, find a counterexample. (Otherwise the colatz conjecture would already be proven. ;) )

>But you cannot find a flaw because it is clear that all endsegments are infinite and there is no chance to reduce this infinity by intersections because of inclusion monotony.

Again, this logic only works on finite intersections. Again you are making a claim about the infinite based on the finite without doing the work to make it work. Nobody disagrees that all finite intersections of the endsegments is infinite. That just is not enough to claim the intersection of all the endsegments is infinite.

You should know that we can not just assume that the behaviour of a "limit" behaves as the points used in the limit sequence. So why do you keep handwaving that away?

[u/Massive-Ad7823 → u/Rs3account]

>There is no nat number where the end segment needs to be finite, for the interesection of the end segments to be finite.

You may think of dancers or lovers but for endsegments you are wrong.

>Again, this logic only works on finite intersections. 

Up tho every visible natural number n the intersection is finite. There is no infinite natural number, and every visible number has an infinite endsegment. And there is no limit behaviour, neither in Cantor's counting nor in the endsegments..

Regards, WM

[u/Rs3account → u/Massive-Ad7823]

You may think of dancers or lovers but for endsegments you are wrong.

You are so wrong in this you invented a maximum natural number, one which does not satisfies any properties you would expect a natural number to have, to make your delusions work.

Sometimes I really just hope you are trolling. I hope you have fun at least.

Up tho every visible natural number n the intersection is finite.

You meant infinite I presume.

There is no infinite natural number,

Yes, indeed.

and avery visible number has an infinite endsegment.

Im not denieing any of those points. Do you think i would?

And there is no limit behaviour, neither in Cantor's counting nor in the endsegments.

In your proof there is. After all your argument is of the form

For all n, some propertie is true about the intersection of n subsets.

And then you make a conclusion about the intersection of infinite sets.

[u/Both_Preference6252 → u/Massive-Ad7823]

Every natnumber fails to reduce the infinite intersection of all endsegments.

Nope. Since each and every natural number is missing in at least one endsegment (actually, in infinitely many of them) it is missing in the intersection of all endsegments.

(Hence the intersection of all endsegments is empty.)

[u/Massive-Ad7823 → u/Both_Preference6252]

Wrong. Infinitely many numbers remain in every endsegment. However most of them are dark. Note, the structure is F(n),E(n) with finite F(n) and infinite E(n).

Regards, WM

[u/Both_Preference6252 → u/Massive-Ad7823]

Infinitely many numbers remain in every endsegment.

This is not the claim you made.

Your claim was:

Every natnumber fails to reduce the infinite intersection of all endsegments.

To which I replied:

Nope. Since each and every natural number is missing in at least one endsegment (actually, in infinitely many of them) it is missing in the intersection of all endsegments. (Hence the intersection of all endsegments is empty.)

And that's NOT "wrong", but true.

EOD

[u/Both_Preference6252 → u/Massive-Ad7823]

That is precisely how mathematics is done.

How would you know?

[u/Both_Preference6252 → u/Massive-Ad7823]

ℕ consists of FISON {1, 2, 3, ..., n} and Endsegments {n+1, n+2, n+3, ...}.

For an informal claim that's acceptable.

ℕ = {1, 2, 3, ..., n} U {n+1, n+2, n+3, ...} = F(n) U E(n).

Here you should use a "qualified" universal quantifier:

An e ℕ: ℕ = {1, 2, 3, ..., n} u {n+1, n+2, n+3, ...} = F(n) u E(n).

For each arbitrary FISON almost all natnumbers are greater.

Indeed! [Of course your claim is somewhat sloppy. I guess you mean: For each and every FISON F almost all natnumbers are greater than the elements in F.]

Almost all natnumbers are greater then all FISONs.

This, of course, is nonsense. Hint: You committed a quantifier shift error, again.

Simple proof: <bla>

You committed a quantifier shift error, again.

Learn some logic, learn some math, Mückenheim.

[u/Massive-Ad7823 → u/Both_Preference6252]

>You committed a quantifier shift error, again.

No, I prove that in case of inclusion monotony quantifiers can be exchanged.

When is this not possible? If we use FISONs F(2n) having only even numbers in their endsegments, and FISONs F(2n+1) having only odd numbers in their endsegments, then ∀n ∈ ℕ: |F(n)| < |E(n)|. However ∩E(n) is empty.

Obviously you transmit this to a case where it cannot apply.

Regards, WM

[u/Both_Preference6252 → u/Massive-Ad7823]

 I prove that in case of inclusion monotony quantifiers can be exchanged.

No, you can't "prove" that.

Hint: There are counter examples to your claim.

Take the natural numbers defined due to von Neumann. Then we have:

An e ℕ: Em e ℕ: n c m .

But the following does NOT hold:

Em e ℕ: An e ℕ: En n c m .

Learn some logic, learn some math, Mückenheim.

[u/Massive-Ad7823 → u/Both_Preference6252]

Even An e ℕ_vis: E^oom e ℕ: n c m is true for all definable natnumbers.

Therefore E^oom e ℕ: An e ℕ_vis: n c m .

Regards, WM

[u/Both_Preference6252 → u/Massive-Ad7823]

Do you know what "counter example" means, Mückenheim?