# Has anyone ever proposed additional axioms?

According to Wikipedia, Godel's incompleteness theorem states:

No consistent system of axioms whose theorems can be listed by an "effective procedure" (essentially, a computer program) is capable of proving all facts about the natural numbers.

This obviously includes our current system. So has anyone proposed any additional axioms that seem credible?

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Do the axiom of choice or continuum hypothesis count? – BlueRaja - Danny Pflughoeft Jul 23 '10 at 15:43
Feferman, Friedman, Maddy & Steel (2000) "Does mathematics need new axioms?" math.ohio-state.edu/~friedman/pdf/DoesMathNeed102601.pdf – Charles Stewart Jul 23 '10 at 20:10
@BlueRaja: Axiom of choice doesn't count - I mean more axioms than the standard, not less. – Casebash Jul 23 '10 at 22:59
Computers cannot "prove" anything. – CogitoErgoCogitoSum Feb 22 '13 at 4:57

## 3 Answers

Set theory is completely full of new axioms, some of them expressing fundamental set principles, and many of them having consequences even in natural number arithmetic that are not provable without them (via consistency strength statements).

In the past fifty years of research in set theory, a major lesson that has been learned is that many many fundamental questions about set theory are simply independent of the usual ZFC axioms. This includes most all of the questions about infinite cardinal arithmetic, but also subtle questons about infinite combinatorics. For example, Is the exponential function (size of power set function) injective on infinite cardinalities? Independent of ZFC. Do Souslin trees exist? Independent of ZFC. Is the size of the smallest dominating family of functions $f:\omega\to\omega$ necessarily $\aleph_1$? Independent of ZFC. Can Lebesgue measure be more than countably additive, when CH fails? Independent of ZFC. There are hundreds of examples.

The response to this phenomenon was naturally to investigate the various hypotheses that go beyond ZFC. So we now know a great deal about what happens when CH holds, or when it fails, or when CH fails but the dominating number is low, and so on. The effect of this is to treat these hypotheses as semi-temporary axioms within a domain of research, which may not be the same usage of axiom that you inquired about, but the effect is the same.

The powerful method of forcing, invented by Cohen in order to prove the consistency of $ZFC+\neg CH$, allows us to show that many other theories are also consistent. This method has given rise to a number of forcing axioms, such as Martin's Axiom, the Proper Forcing Axiom, Martin's Maximum and many others.

A much larger and very interesting class of axioms are provided by the large cardinal hierarchy, involving such notions as inaccessible cardinals, measurable cardinals, Woodin cardinals and so on. A curious feature of these strong axioms of infinity is that they imply certain highly regular features to occur among the projective sets of reals. Thus, although the axioms themselves seem to have nothing to do with sets of reals, they imply very nice properties for the sets of reals that we can easily define. Many set theorists take this as some kind of positive evidence for their truth. Meanwhile, the large cardinal hypotheses are intensely studied as a fundamental research effort to understand the nature of mathematical infinity. But you cannot do this without assuming that these cardinals exist, since this is provably not provable in ZFC, if consistent, and so these large cardinal hypotheses constitute new axioms.

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I've always thought it would be really cool if P versus NP turned out to be unprovable in ZFC (especially since this is the case with respect to one decidable oracle). – Akhil Mathew Aug 7 '10 at 1:04

Adding additional axioms would make more truths provable. But it wouldn't make all truths provable (unless the axiom was inconsistent with the already given ones, in which case all falsehoods would also be provable too).

So adding additional axioms isn't going to help make all the truths provable. I guess you could just add, say, Goldbach's conjecture and the Riemann hypothesis as extra axioms and carry on doing mathematics-PLUS! but why would you want to?

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Seamus: I doubt that Goldbach's conjecture or the Riemann hypothesis would make good axioms. The point of an axiom is to be intuitive. The continuum hypothesis seems to be a better example though – Casebash Jul 23 '10 at 23:07
@Casebash: It may very well turn out that the Riemann hypothesis is independent of ZFC, at which point, to keep our sanity, it will likely become an axiom (or maybe someone will find a simpler axiom which allows us to prove it). – BlueRaja - Danny Pflughoeft Jul 23 '10 at 23:32
Adding axioms makes more theorems provable. They need not be though of as truths. You can add either the Axiom of Choice or its denial to standard set theory, extending the set of theorems in disjoint ways. On one natural interpretation of that, only one of those two sets of novel theorems is a set of truths. – vanden Aug 6 '10 at 19:49
I think it is highly unlikely that Goldbach or RH are independent of ZFC. They are surely either true of false, although we don't know which yet. Eventually we will; in the meantime, it would be a mistake to prejudge the issue. (Although of course, people frequently prove results contingent on RH, say; but such results are rightly regarded as contingent.) – Matt E Aug 7 '10 at 5:57
Also, one could indeed add enough axioms to make all truths provable. E.g. one can consider the standard model of the natural numbers (the one consisting of {1,2,3,...}), and take the set of all true theorems for this model. This will be consistent and contain all truths about the natural numbers. Godel's result simply shows that it is not recursively describable; that is not the same as saying that it does not exist. – Matt E Aug 7 '10 at 5:59

Sure, the continuum hypothesis. We know, thanks to the work of Godel and Paul Cohen, that this is a legitimate axiom, in that it doesn't destroy consistency: you can't disprove it (or prove it) using Zermelo-Frank set theory.

People do actually use the continuum hypothesis in model theory, for instance, where there are several results that depend on it. For an article about its uses, cf. the Tricki page.

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Cohen showed that neither Choice nor the Continuum Hypothesis could be proved in ZF. It was earlier work of Goedel's that showed that they cannot be disproved in ZF. – vanden Aug 6 '10 at 19:47
Corrected, thanks. – Akhil Mathew Aug 6 '10 at 20:00
My understanding is that Godel (who I believe was a set-theoretic Platonist), among others, believed that CH was false. In other words, while it may give a consistent theory when adjoined to ZF, the resulting theory may not actually describe true sets. – Matt E Aug 7 '10 at 6:03