chinglu hates the Axiom of Infinity

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

  1. chinglu Valued Senior Member

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    Anyone know a justification that the axiom of infinity describes a set that contains all natural numbers?
     
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  3. PhysBang Valued Senior Member

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    By definition of natural numbers?

    Seriously, that's the answer: the axiom says that there is a set that contains a certain kind of thing and the natural numbers are defined to be those sorts of things.
     
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  5. chinglu Valued Senior Member

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    like from here or kunen's book. remember set theory is described with the language of first order logic with epsilon as the element of predicate.

    https://en.wikipedia.org/wiki/Axiom_of_infinity
     
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  7. PhysBang Valued Senior Member

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    If you know the answer, then why ask?
     
  8. paddoboy Valued Senior Member

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    The mind boggles!

    Please Register or Log in to view the hidden image!

     
  9. chinglu Valued Senior Member

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    i simply indicated that your answer was a fail and not relevant to the question.

    has nothing to do with my question.
     
  10. rpenner Fully Wired Valued Senior Member

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    There is not just one way to express a viable "Axiom of Infinity", nevertheless we refer to it as "the" Axiom of Infinity -- the definite article indicates they are all equivalent.

    S is a non-specific infinite set. s, o, O are sets. ω is the class containing the all the finite ordinal numbers. V is the universe of all sets.

    ax-inf0 : \( \vdash \exists S \left( s \in S \wedge \forall o \left( o \in S \to \exists O \left( o \in O \wedge O \in S \right) \right) \right)\) It is true that there exists at least one nonempty set S (which at least includes set s) such that for every set o in S, S also contains at least one set O which in turn contains o.
    ax-zfinf : \( \vdash \exists S \left( o \in S \wedge \forall o \left( o \in S \to \exists O \left( o \in O \wedge O \in S \right) \right) \right)\) It is true that there exists at least one nonempty set S such that for every set o in S, S also contains at least one set O which in turn contains o.
    ax-inf1 : \( \vdash \exists S \left( S \neq \emptyset \wedge \forall o \left( o \in S \to \exists O \left( o \in O \wedge O \in S \right) \right) \right)\) It is true that there exists at least one nonempty set S such that for every set o in S, S also contains at least one set O which in turn contains o.
    ax-inf2 : \( \vdash \exists S \left( S \neq \emptyset \wedge S\subseteq \cup S \right) \) It is true that there exists at least one nonempty set S such that S is a subset or equal to the union of all members of S.
    ax-inf3 : \( \vdash \omega \in V \) It is true that the class of all finite ordinal numbers is a member of the class of all sets.
    ax-inf5 : \( \vdash \exists S \; S \subset \cup S \) It is true that there exists a set S which is a proper subset of the union of its elements.

    Logically, any of these follows from any of the others. Thus we have a graph of logical implications without any loose leafs.

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    Output of echo 'digraph impl { node [label="ax-\N"]; inf5 -> inf3 -> inf0 -> zfinf -> inf1 -> inf2 -> inf0 ; zfinf -> inf2 ; inf2 -> inf3 -> inf5 ; }' | circo -Tgif > impl.gif
    The above has been adapted from metamath.org mostly because I don't want to open a textbook.
     
    Last edited: Jun 15, 2016
  11. PhysBang Valued Senior Member

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    I see that you never actually looked at the definitions involved.
     
  12. Dinosaur Rational Skeptic Valued Senior Member

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    Cantor wrote an essay which did an excellent job of dealing with infinity.
     
    chinglu likes this.
  13. chinglu Valued Senior Member

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    So, anyway Kunen stated the the same definition as the wiki post I gave.

    can you justify it yes or no.
     
  14. chinglu Valued Senior Member

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    can you justify the axiom or not?

    Remember, axioms must be obvious.
     
  15. chinglu Valued Senior Member

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    Prove that.
     
  16. arfa brane call me arf Valued Senior Member

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    I can justify the existence of an infinite number of decimal places very easily.

    But I can't write them all down. But I don't have to, because I have to accept that say, 1 divided by 3 has an infinite sequence of 3s after the decimal point.
    It's something so obvious it must be axiomatic.

    Moreover, dividing 1 by 3 establishes the existence of a natural number of decimal places after the ".", there is no largest natural number nor is there a final decimal place in an infinite sequence of them.

    And logically, numbers with finite decimal expansions can be extended with zeros, since you can add any number of zeros you can add an infinite sequence of them.
     
    Last edited: Jun 18, 2016
  17. rpenner Fully Wired Valued Senior Member

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    That's not true.

    Axioms by their very nature are unprovable, so it does not follow (from anything) that they must be obvious.

    It follows from the definition of the natural numbers that there is no largest natural number, but it is not obvious that there is a set of ALL natural numbers. No amount of rational discussion will incline one to adopt the axiom of infinity as either true or false. It's possible to do mathematics while assuming the axiom of infinity is false, but such mathematics doesn't contain the real numbers or other many other familiar elements.

    However, when doing ZFC Set theory, one assumes its axioms (including the axiom of infinity) are all true, so you're stuck with it.
     
  18. PhysBang Valued Senior Member

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    In the interests of some clarity (I hope), not to just be pedantic:
    This is something that is not directly related to the OP. Still, what you can justify is the need to indefinitely extend certain representations of certain specific real or rational numbers.
    Proving the indefinite extension is something that one proves in computational theory, it is not usually taken to be obvious.

    And, again, the proof establishes the indefinite extension of a particular representation of a number, not the existence of a particular set. The axiom of infinity sets up mathematical rules through establishing a certain metaphysics of sets; it is for this reason that some people find it distasteful.
    By the definitions, 1 divided by 3 is not a natural number, it is a rational number (and a real number). I bring this up because it does not directly bear on the OP.
    This in principle practice assumes something like the axiom of infinity, since it assumes the metaphysics of countable infinity and a relationship between the existence of a set of countably infinite numbers and theoretical decimal representations. As such it seems to assume the outcome of what the OP wants guidance on.

    The axiom of infinity is a way of establishing by fiat that one infinite set exists and then using that set to build other infinite sets. This metaphysical approach is viewed by many as suspect or at least inelegant. (For some mathematicians and logicians, the latter is far worse!)
     
  19. arfa brane call me arf Valued Senior Member

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    The computation that one uses to prove there is an indefinite extension of decimal places in the decimal expansion of 1/3, each containing the same number, is called division. It's algorithmic.

    If you use division algorithmically, it's obvious that dividing 1 by 3 (as an example) always leaves the same remainder. So the algorithm can continue indefinitely.
    So one must accept that there is no last remainder. The "justification" (viz the OP) for accepting something obvious, is that it's axiomatic. There is a first decimal place, a second, third, etc. So there is a direct map from the natural numbers to the sequence of decimal places.

    One of Euclid's geometric axioms is that any two points in the plane can be connected by a line. This is not controversial, it's accepted without (usually) questioning why it's true.
     
  20. PhysBang Valued Senior Member

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    Sure. And an algorithm is something that can be done over and over. It is not something that necessarily is done over and over, in some sense.
    That there is no last remainder in the decimal system is not itself enough to establish that there actually is an infinite set that is the natural numbers. The axiom of infinity just says that this set actually exists. Using that, we can then say all kinds of other stuff, like there actually is a set with all the possible digits of the decimal representation of 1/3. But the existence of the algorithm alone is not enough because the algorithm just lets us keep going--it doesn't say that these steps have already been completed in some way. Hence the desire to create something like the axiom of infinity.

    I'm not sure exactly what the OP is asking, but the justification as to why the axiom of infinity describes a set that contains all natural numbers is because, usually, the axiom just includes the definition of natural numbers and says that they are all in a single set.
     
  21. arfa brane call me arf Valued Senior Member

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    5,751
    I think, though I could be wrong, that the axiom that establishes the infinity of the natural numbers is that there is no greatest natural number. Taking that with the obvious fact that there is a smallest natural number (1), the division algorithm does in fact say the same thing--if there are n division 'steps' and n remainders, there is an n+1th step and n+1th remainder.



    It's easy to write a program that never halts, whether it also outputs anything other than blanks. Proving the program never halts by running it will take forever. Nonetheless the program will never halt, and this can be proven logically.

    That isn't the halting problem, which is about proving any program will halt, not a specific program.
     
  22. PhysBang Valued Senior Member

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    In this case, you are kinda wrong. The axiom is supposed to capture that there is no greatest natural number in set theory, so it just adds the set to set theory.

    http://mathworld.wolfram.com/AxiomofInfinity.html

    (And, sadly, https://en.wikipedia.org/wiki/Axiom_of_infinity )
    But this is different from a program that has actually run forever. This is an important metaphysical distinction.
     
  23. arfa brane call me arf Valued Senior Member

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    5,751
    Yeah, "justify an axiom" is an oxymoron because an axiom is an accepted truth, philosophically. You accept that a line exists between two points before you do any geometry or surveying, for that matter. I think the OP thought there was some kind of gotcha, but I don't see it.
    The axiom is supposed to capture that [what I said] in set theory, expands on what I said exactly how?
     

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