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Goldstern
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The reflection principle is a theorem scheme; each of its instances is provable in ZFC.

The following proof works entirely in ZFC:

  1. Assume Con(ZFC) together with not-Con(ZFC+non-CH) and aim for a contradiction.

  2. By assumption plus completeness theorem, there is a model (M,E) of ZFC. M may be nonstandard in the sense that its ordinals (even its natural numbers) are not well-founded.

  3. By compactness, there is some finite sub theory ZFC* of ZFC such that ZFC*+non-CH is inconsistent.

  4. Find a (slightly larger) finite sub theory ZFC** of ZFC such that ZFC** proves all theorems you need about forcing, including that ZFC* plus non-CH holds in the extension.

  5. (M,E) satisfies reflection for ZFC**ZFC** , so there is some x in M which M thinks is a ctm (countable transitive model) of ZFC**ZFC** . So M can find a Cohen extension y of x which satisfies ZFC* plus non-CH.

  6. By the easy versiondirection of the completeness theorem, M therefore thinks that ZFC*+non-CH is consistent.

  7. But then this theory is really consistent. (BecausesBecause any proof of an inconsistency would be coded by a natural number, which is represented in M.)

Btw, I think the sketch I just gave is not ideal since it seems to rely on the existence of infinite sets. The implication from Con(ZFC) to Con(ZFC+non-CH) can really be shown in ZF minus infinity, or even in a weak version of Peano arithmetic.

The reflection principle is a theorem scheme; each of its instances is provable in ZFC.

The following proof works entirely in ZFC:

  1. Assume Con(ZFC) together with not-Con(ZFC+non-CH) and aim for a contradiction.

  2. By assumption plus completeness theorem, there is a model (M,E) of ZFC. M may be nonstandard in the sense that its ordinals (even its natural numbers) are not well-founded.

  3. By compactness, there is some finite sub theory ZFC* of ZFC such that ZFC*+non-CH is inconsistent.

  4. Find a (slightly larger) finite sub theory ZFC** of ZFC such that ZFC** proves all theorems you need about forcing, including that ZFC* plus non-CH holds in the extension.

  5. (M,E) satisfies reflection for ZFC**, so there is some x in M which M thinks is a ctm (countable transitive model) of ZFC**. So M can find a Cohen extension y of x which satisfies ZFC* plus non-CH.

  6. By the easy version of the completeness theorem, M therefore thinks that ZFC*+non-CH is consistent.

  7. But then this theory is really consistent. (Becauses any proof of an inconsistency would be coded by a natural number, which is represented in M.)

Btw, I think the sketch I just gave is not ideal since it seems to rely on the existence of infinite sets. The implication from Con(ZFC) to Con(ZFC+non-CH) can really be shown in ZF minus infinity, or even in a weak version of Peano arithmetic.

The reflection principle is a theorem scheme; each of its instances is provable in ZFC.

The following proof works entirely in ZFC:

  1. Assume Con(ZFC) together with not-Con(ZFC+non-CH) and aim for a contradiction.

  2. By assumption plus completeness theorem, there is a model (M,E) of ZFC. M may be nonstandard in the sense that its ordinals (even its natural numbers) are not well-founded.

  3. By compactness, there is some finite sub theory ZFC* of ZFC such that ZFC*+non-CH is inconsistent.

  4. Find a (slightly larger) finite sub theory ZFC** of ZFC such that ZFC** proves all theorems you need about forcing, including that ZFC* plus non-CH holds in the extension.

  5. (M,E) satisfies reflection for ZFC** , so there is some x in M which M thinks is a ctm (countable transitive model) of ZFC** . So M can find a Cohen extension y of x which satisfies ZFC* plus non-CH.

  6. By the easy direction of the completeness theorem, M therefore thinks that ZFC*+non-CH is consistent.

  7. But then this theory is really consistent. (Because any proof of an inconsistency would be coded by a natural number, which is represented in M.)

Btw, I think the sketch I just gave is not ideal since it seems to rely on the existence of infinite sets. The implication from Con(ZFC) to Con(ZFC+non-CH) can really be shown in ZF minus infinity, or even in a weak version of Peano arithmetic.

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Goldstern
  • 14k
  • 1
  • 47
  • 71

The reflection principle is a theorem scheme; each of its instances is provable in ZFC.

The following proof works entirely in ZFC:

  1. Assume Con(ZFC) together with not-Con(ZFC+non-CH) and aim for a contradiction.

  2. By assumption plus completeness theorem, there is a model (M,E) of ZFC. M may be nonstandard in the sense that its ordinals (even its natural numbers) are not well-founded.

  3. By compactness, there is some finite sub theory ZFC* of ZFC such that ZFC*+non-CH is inconsistent.

  4. Find a (slightly larger) finite sub theory ZFC** of ZFC such that ZFC** proves all theorems you need about forcing, including that ZFC* plus non-CH holds in the extension.

  5. (M,E) satisfies reflection for ZFC**, so there is some x in M which M thinks is a ctm (countable transitive model) of ZFC**. So M can find a Cohen extension y of x which satisfies ZFC* plus non-CH.

  6. By the easy version of the completeness theorem, M therefore thinks that ZFC*+non-CH is consistent.

  7. But then this theory is really consistent. (Becauses any proof of an inconsistency would be coded by a natural number, which is represented in M.)

Btw, I think the sketch I just gave is not ideal since it seems to rely on the existence of infinite sets. The implication from Con(ZFC) to Con(ZFC+non-CH) can really be shown in ZF minus infinity, or even in a weak version of Peano arithmetic.