Can Countable Axioms Limit The Number of Theorems About Real Numbers?

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Countable axioms limit the number of theorems about real numbers to a countable set, as proofs derived from these axioms are finite sequences of statements. Even with an infinite number of axioms, the derived theorems remain countable, meaning no uncountable number of statements can be produced from a countable set of axioms. The discussion highlights the distinction between explicit and implicit definitions of axioms, emphasizing that explicit definitions lead to a clearer understanding of derived statements. The conversation also touches on the idea of defining an order relation on proofs, suggesting a theoretical exploration of proof limits. Ultimately, the consensus is that countable axioms do not expand the scope of mathematical knowledge beyond countability.
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If I have an \aleph_0 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.
 
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cragar said:
If I have an \aleph_0 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.

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.
 
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.
 
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.
 
cragar said:
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.

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.
 
Bacle2 said:
good question, BTW.

Yes!

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".
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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