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Question about axioms

  1. Aug 21, 2012 #1
    If I have an [itex] \aleph_0 [/itex] number of axioms, does that put a limit on the number of theorems I can have about the real numbers.
    The number of theorems that we could have is countable. I was just wondering what we might be able to say about how much we could know about math.
  2. jcsd
  3. Aug 21, 2012 #2
    A proof of a theorem is a finite sequence of statements, each one of which is either an axiom or is derived from previous lines of the proof. There are countably many proofs of length 1; countably many proofs of length 2, dot dot dot.

    The union of the proofs of length n as n ranges over 1, 2, ... is a countable union of countable sets, so it's countable.

    In other words, a countably infinite number of axioms doesn't buy you any more math than a finite set of axioms.
  4. Aug 21, 2012 #3
    ok. So if I start with a finite number of axioms I could derive a countably infinite number of statements from that. And after I did that those statements are basically my axioms.
    Ok I see what you are saying. So there is no way to derive an uncountable number of statements from a countable set.
  5. Aug 22, 2012 #4


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    This is a very interesting question.

    One thing I have to ask is whether the axioms are explicitly defined or not.

    It may sound like a stupid question, but if you can define an axiom non-explicitly then you might have a different kind of case to work with as opposed to defining them all explicitly.

    So what I mean is that by explicit you have an explicit definition for the axioms and then as SteveL27 said, you generate all possible statements as derived from those axioms (i.e. the rest of the axioms are 'unpacked' from the definition of the minimal set).

    In this context the language used to define the axioms are explicit since the axiomatic definitions can not change. In an implicit context, this doesn't hold.
  6. Aug 22, 2012 #5


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    An uncountable number of statements, yes, but not an infinite number of proofs, if

    you consider a proof to be a finite collection of statements, as SteveL pointed out. Sorry if this is what you

    meant--good question, BTW.
  7. Aug 22, 2012 #6

    Stephen Tashi

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    If we let our minds race we can daydream about Dedekind cuts for proofs. We'd need to define an order relation on proofs and have an axiom that says "The limit of any bounded monotonic sequence of finite proofs is a proof".
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