# chinglu hates the Axiom of Infinity

Discussion in 'Pseudoscience' started by chinglu, Jun 14, 2016.

1. ### chingluValued Senior Member

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• Please don't insult other members
circus clown.

"infinity isn't a complete collection"
https://en.wikipedia.org/wiki/Controversy_over_Cantor's_theory

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3. ### PhysBangValued Senior Member

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So, you just provided the context required to make that statement mean anything, while pretending not to. This is why people think poorly of you and restrict your threads to pseudo science.

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5. ### arfa branecall me arfValued Senior Member

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Well, you've introduced a concept that goes back to Aristotle at least.
Is there an actual infinity of digits in the expansion of 1/3 (or any other rational number with an infinite expansion)? Or, is the division algorithm always incomplete (never halting)?

Does anyone care all that much these days? My response to your "show me a completed infinite collection", is that infinity can't be completed other than in the sense of being axiomatically completed--you can write an indefinite number of programs, all finite (complete) in length that will never halt, that will run infinitely. This is despite lack of resources like space and time. The natural numbers are infinite, and axiomatically 'complete', but not in the sense of having some number that represents that completion.

PhysBang likes this.

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7. ### arfa branecall me arfValued Senior Member

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And let's think some more about a program that never halts, and what that might mean in terms of using up resources which can only be finite (unless the universe is actually infinite).

Staying with the example of 1/3, the representation of this number just written, is complete because it uses a finite number of symbols, a finite amount of space, and it took a finite amount of time to compute and output.

If I want to represent the infinite decimal expansion, that's an unending sequence of 3s I have to write (or get a machine to write for me). If I relax the conditions on the output and decide I can write the 3s anywhere, I can write them all in the "same place" (using up finite resources), and it will still take forever to write them all.

I can also write a program that calculates all the digits in the expansion one after the other, which doesn't output anything, and it will still take forever. It won't need space for the output so the universe can be finite in extent (whew!) . . .

8. ### arfa branecall me arfValued Senior Member

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I think what I'm saying there is I can take care of the objection that infinite space is needed to write an actually infinite sequence, by having a program use the same space (or no space) to write the sequence. But then it isn't a sequence, at least, not in space. But it is still a sequence in time.

Ok, but to be infinite, the universe must be infinite at least in time. If it is, I can start a program that calculates an infinite sequence anytime and it will still run forever (assuming I can also build a machine that lasts forever).

Numbers like 1,2,3 do last forever (and they already have lasted that long, numbers really can't be said to have a history although humans do have one), they don't occupy space, nor does the infinite set of them.

Last edited: Jun 25, 2016
9. ### rpennerFully WiredValued Senior Member

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The mathematician in me thinks that's the wrong way to go about thinking about the infinite.
Something that must be created or verified element-by-element implies a process that will literally take forever. But if you presume the infinite set into existence all at once, there is no such issue.

This allows us to reason about the infinite even though we never have the resources to construct or enumerate the infinite.

10. ### arfa branecall me arfValued Senior Member

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I think that's what I do in the last sentence; "presume the existence of numbers without writing any" I thought was something like what I say above.

11. ### PhysBangValued Senior Member

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Programs do not calculate: people and, in a metaphorical sense, machines calculate. A machine runs a program only if it produces the appropriate output.

12. ### arfa branecall me arfValued Senior Member

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Machines which are like computers, calculate if they run programs. Computers calculate where the "next" instruction is for example. Testing or setting logical variables is a calculation.

Given that a machine can be a set of instructions--a program--running on another machine, I don't personally see why there's any problem with how I used the word "calculate".

But what should I do if asked to write a program that calculates the first n digits of pi? I guess I should say I can't, sorry, the English language won't let me do that.

God, sarcasm is such a hassle sometimes.

Last edited: Jun 26, 2016
13. ### PhysBangValued Senior Member

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Sure, they calculate in a metaphorical sense. Calculation is a mental activities and machines do not think. We can rig machines to run in a way that they end up in a state that is representative of the end of a calculation because of the neat features of calculation. But a machine does something physical, while when we calculate we do something physical that is also mental.

Mostly, this difference is of no concern. However, if we are talking about the foundations of mathematics, it really is important.

14. ### arfa branecall me arfValued Senior Member

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I understand the word calculate (and calculus), are from the Latin for "small stone or pebble".
A lot of dictionaries will tell you that calculate and compute are synonyms. Whereas calculate usually means a numerical process and a numerical result, compute may not mean that (since you could get a "yes/no" result instead).

I've never seen a problem (and I have a degree in CS). So what is important when talking about foundations?

15. ### PhysBangValued Senior Member

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I don't see any difference between calculate and compute, in the mathematical sense, since we can represent a yes-or-no result numerically.

If we are speaking of the foundations of mathematics, then we might want to ask ourselves questions about the status of infinities. For example, we can do a lot of constructive mathematics with countable infinities, but if we want to do all of calculus, then we need uncountable infinities. Should we accept all of calculus? We don't need all of calculus, it turns out: we can do all sorts of physics without all of it.

But constructive calculus looks weird: if a smooth function has values in its range above zero and below zero, you can in general only prove that you can find values in its range arbitrarily close to zero, not actually at zero. Constructive mathematics requires that you actually demonstrate that there is a specific value in the domain that the function takes to zero if you want to prove that.

Constructive methods limit us to arbitrary close statements, but so too does physics, at least since all our applications and experiments have some error to them.

Constructive methods use axioms; they can even handle the axiom of infinity (at least in some forms). Computation (as a mathematical field) is constructive.

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prove that.

17. ### chingluValued Senior Member

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Does not work.

If you have to do that to get an infinite set, then you must do that for all infinite sets by hypothesis. Hence, any infinite collection must be postulated to exists.

exactly what kind of math is that.

18. ### chingluValued Senior Member

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I will now teach those who do not understand theoretical math how a set of all natural numbers is justified.

Now, this is from Kunen since I can see none here understand his book.

Axiom of Infinity- $\exists x(0 \in x \land \forall y \in x(S(y) \in x))$

(Kunen) - INF - If a set x satisfies the axiom of infinity then x contains every natural number.
Proof.
Let the predicate $NN(y)$ mean y is a natural number. Assume $\exists n (NN(n) \land n \notin x )$. Form the set $X=\{y:NN(y) \land y\leq n \land y \notin x\}$. X is non-empty by assumption so choose the least $n' \in X$ then $n' \notin x$. Since $0 \in x$ by INF then $n' \neq 0$. So, there is some m' such that $NN(m')$ and $S(m')=n'$. Hence, $m'\in x$ since n' is the least element in X. But, then by INF $S(m')=n'\in x$ which contradicts $n' \notin x$. Thus, $\neg\exists n (NN(n) \land n \notin x )$ and then $\forall n (NN(n) \to n \in x ).$

19. ### arfa branecall me arfValued Senior Member

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Can you translate what the first expression says, into English? That could give me more confidence that a) you understand Kunen's book, and b) you can teach anything to anyone.

Or not.

20. ### PhysBangValued Senior Member

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What kind of math? The kind of math we have and have known that we have for almost 100 years now.

You know, chinglu, there are actually textbooks about mathematics that you can read and classes that you can take.

21. ### PhysBangValued Senior Member

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You are demonstrating that you don't understand his book.
Ta da! You just showed a hypothetical proof.

22. ### chingluValued Senior Member

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be more specific.

then show why this is not kunen's argument if that is your claim.

23. ### chingluValued Senior Member

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so, are you claiming the above is not in Kunen's book?