Undecidable statement made into axiom?

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Discussion Overview

The discussion revolves around the implications of adding undecidable statements, specifically the Continuum Hypothesis (CH), as axioms within the Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC). Participants explore whether adding such statements can maintain consistency in the system and the ongoing debates surrounding their truth values.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants propose that if a statement is undecidable in an axiomatic system, it can be added as an axiom without leading to contradictions, assuming the original system is consistent.
  • There is a suggestion that adding the negation of the CH to ZFC could allow for the existence of a cardinal x such that aleph_0 < x < c.
  • Others express confusion about why the CH remains an open question if it can be added as an axiom, questioning the implications of its independence from ZFC.
  • Some participants assert that the independence of the CH from ZFC implies that its negation is also independent of ZFC, although this is met with some skepticism regarding the status of the CH itself.
  • There are discussions about the usefulness of classifying other statements in relation to the CH and the implications of assuming or negating it.
  • One participant notes that the article they were reading was discussing the Generalized Continuum Hypothesis (GCH) rather than the CH, which led to some confusion.

Areas of Agreement / Disagreement

Participants express differing views on the implications of adding undecidable statements as axioms, with some agreeing on the consistency of such actions while others question the ongoing relevance of the CH and its negation. The discussion remains unresolved regarding the broader implications of these statements within ZFC.

Contextual Notes

There are references to the independence of the CH and its negation from ZFC, but participants express uncertainty about the exact status of these statements and their implications. The discussion also highlights potential misunderstandings regarding the distinction between the CH and the GCH.

gravenewworld
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i am a little bit confused on 1 thing. can a statement that is undecidable in a axiomatic system just be added as a axiom to the original system and never lead to a contridiction?

for example godel and cohen showed that the continuum hypothesis is independent of ZFC. does this mean then that if we add the CH or its negation as an axiom to ZFC we will still have a system that is still consistent (assuming ZFC is already consistent)? Does this mean then that if I add the negation of the CH i can show that there is some cardinal x such that aleph_0<x<c?

this is the wiki article that is confusing me-
http://en.wikipedia.org/wiki/Axiom_of_choice#Independence_of_AC

quote from article: "By work of Kurt Gödel and Paul Cohen, the axiom of choice is logically independent of the other axioms of Zermelo-Fraenkel set theory (ZF). This means that neither it nor its negation can be proven to be true in ZF. Consequently, assuming the axiom of choice, or its negation, will never lead to a contradiction that could not be obtained without that assumption."

so if the CH( or ~CH) can be added to ZFC(since it was proven independent just like AC), then why is it still unsolved?
 
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i am a little bit confused on 1 thing. can a statement that is undecidable in a axiomatic system just be added as a axiom to the original system and never lead to a contridiction?
You are correct.

If \Phi is some collection of statements, and \phi is a statement that is "undecidable" in \mathcal{TH}(\Phi)... that is, neither \phi nor \neg \phi follow from \Phi... then \Phi \cup \{ \phi \} is a consistent collection of statements. (And so is \Phi \cup \{ \neg \phi \})


(P.S. anybody know how I can get a nice, fancy lower-case script letter? If I try to turn Th into the mathcal font, I get \mathcal{Th})
 
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i think it is decidable, in fact i have decided the answer is no.
 
Hurkyl said:
You are correct.

If \Phi is some collection of statements, and \phi is a statement that is "undecidable" in \mathcal{TH}(\Phi)... that is, neither \phi nor \neg \phi follow from \Phi... then \Phi \cup \{ \phi \} is a consistent collection of statements. (And so is \Phi \cup \{ \neg \phi \})(P.S. anybody know how I can get a nice, fancy lower-case script letter? If I try to turn Th into the mathcal font, I get \mathcal{Th})
Ok, I still don't understand why then people are still trying to prove the falsity of the continuum hypothesis. If the continuum hypothesis is independent of ZFC, then it can be added as an axiom. But if it is an axiom shouldn't it be universally true, why would people still bother trying to negate it?
 
If the negation of the continuum hypothesis is independent of ZF, then it could be added as an axiom.

One might not want to assume the CH. Just as people might not want to assume the AC.

And then, one might want to classify other statements in terms of how they relate to the CH. In other words, one might find it useful to know that P=>~CH. Among other things that would prove that P is independent of ZF.

(You sure the CH is independent of ZFC? I know it is independent of ZF, but I can't remember ZFC for sure)
 
Hurkyl said:
If the negation of the continuum hypothesis is independent of ZF, then it could be added as an axiom.

One might not want to assume the CH. Just as people might not want to assume the AC.

And then, one might want to classify other statements in terms of how they relate to the CH. In other words, one might find it useful to know that P=>~CH. Among other things that would prove that P is independent of ZF.

(You sure the CH is independent of ZFC? I know it is independent of ZF, but I can't remember ZFC for sure)


But doesn't the fact that the CH is independent of ZFC already imply that its negation is independent of ZFC (or is the article i am reading simply wrong)? I also am pretty sure that CH has already been proven independent of ZFC.
 
gravenewworld said:
But doesn't the fact that the CH is independent of ZFC already imply that its negation is independent of ZFC
Yes. Incidentally, that wiki article is talking about the GCH, not the CH.
 
Hurkyl said:
Yes. Incidentally, that wiki article is talking about the GCH, not the CH.


I actually was reading that article and from what it said about the AC, I just made an analogy to the CH (cohen and godel showed that CH is independent of ZFC). So if we add ~CH to ZFC, it will be consistent. Doesn't this mean now that we can find a cardinal call it x such that aleph0<x<c?
 
Yes. ZFC~CH implies the existence of such a cardinal.
 
  • #10
alright thanks a bunch. i think i am out of questions now.
 

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