Persuasive proposal versus actual proof of independence of the parallel postulate

[PseuTonym]

To actually prove the independence of the parallel postulate from the other premises of Euclidean geometry, we need some kind of model theory.

Unfortunately, the usual approach to model theory is to represent a predicate by means of a set, to represent a binary relation by means of a set of ordered pairs, etc. What motivates me to use the word “unfortunate” is Russell’s paradox.

Define the predicate P(x) to mean that it is not the case that x is an element of x.
Corresponding to P(x), there is no set. However, P(x) is a perfectly good predicate. So, it seems that we cannot use all possible sets to represent all possible predicates. We have exhibited an example of a predicate that cannot be represented (using the usual method) via a set. Perhaps there is some other way of using a set to identify a predicate, but the usual method doesn’t seem to work.

What is an approach to model theory that was developed after Russell’s paradox was discovered, and that acknowledges the difficulty that is raised by Russell’s paradox?

 

[Rhubarb]

Russell developed “Ramified Type Theory” (If I’m recalling correctly) to combat his paradox.

I know there are efforts to continue Russell’s logicist project - providing a basis for arithmetic on logic alone. But I don’t know much about the modern arguments.

Zermelo’s set theory avoids the paradox too, but I couldn’t explain how. I just know that’s the general consensus.

 

[PseuTonym]

Zermelo’s set theory avoids the paradox too

There is a difference between a theorem within a formal system, such as ZF, and a meta-mathematical result.

Recall that the second part of Godel’s incompleteness theorem is the unprovability of consistency. To prove the unprovability of consistency, we need to assume that the system is consistent. If the system is inconsistent, then there is a formal proof of every well-formed formula, including well-formed formulas that we interpret as asserting that the system is consistent. In other words, if the system is inconsistent, then a formal proof based on the system doesn’t actually tell us anything about the well-formed formula that was proved.

When people repeat the claim that replacing the parallel postulate with various alternative postulates gives us alternative systems of geometry that are consistent and legitimate, they are talking about a meta-mathematical result. To say “this system is consistent” is to say something about that system.

Kurt Gödel showed in 1940 that the continuum hypothesis (CH for short) cannot be disproved from the standard Zermelo–Fraenkel set theory (ZF), even if the axiom of choice is adopted (ZFC) (Gödel (1940)). Paul Cohen showed in 1963 that CH cannot be proven from those same axioms either (Cohen (1963) & Cohen (1964)). Hence, CH is independent of ZFC. Both of these results assume that the Zermelo–Fraenkel axioms are consistent; this assumption is widely believed to be true.

Link:
en.wikipedia.org/wiki/Continuum_hypothesis

If the axioms of ZF are true, then they are consistent with each other. It’s interesting that the article refrains from asserting that the axioms of ZF are widely believed to be true.
Why build a large body of theorems on the basis of ZF if the axioms of ZF aren’t believed to be true?

If somebody publishes a proof of CH from the axioms of ZF, then we will have a new theorem in ZF. If somebody publishes a refutation of CH based on the axioms of ZF, then we will have a new theorem in ZF. That would be regular mathematics, at least in this era when ZF is a kind of dogma or gamble that is accepted.

What Cohen and Godel did with CH was meta-mathematics.

 

[Rhubarb]

I couldn’t speak on Zermelo as my only foray into philosophy or mathematics was a survey of the history of philosophy of mathematics. Also, my knowledge of actual mathematics extends to a C in calculus twelve years ago. I can only report that I’ve been told that Zermelo’s set theory avoids Russell’s paradox.

The type theory that Russell suggested to beat his own paradox suggests that sets have types. I’m not up on the intricacies but the jist is that sets of a type cannot contain sets of an equal type. There is probably a lot more to that, but as I said. I no maths so gud. The trick is to avoid the recursive definition that “a set of all sets” is an example of.

There are also methods of basing arithmetic that don’t rely on pure logic - my Phil. Maths professor was into Constructivism, herself.

 

[PseuTonym]

The simplest idea is to consider the possibility that every relation is linked to a set of ordered pairs in the following way:
For every binary relation R, there exists a set m such that for all x and y,
(x R y if and only if (x,y) is an element of m).

Here we are thinking of a binary relation R such as the “is an element of” relation.

In particular, we obtain the following via plugging in the relation R that is the negation of the “is an element of” relation…

There exists a set m such that, for all x,
(x is not an element of x) if and only if (x,x) is an element of m.

We have replaced the usual native assumption by substituting the ordered-pair (x,x) for the variable x. If it is possible to deduce a contradiction, then it is at least different from the one that usually arises under the name “Russell’s paradox.”

The next step would be to generalize what is presented above. Given an arbitrary predicate P(x), we obviously don’t get the existence of m such that for all x, P(x) if and only if (x,x) is an element of m. Instead, the ordered-pair (x,x) will be replaced with something that depends on the formula P(x).

 

[PseuTonym]

In particular, we obtain the following via plugging in the relation R that is the negation of the “is an element of” relation…

I apologize for the above error. The above doesn’t accurately describe what relation was plugged in. It seems to be a good idea to use symbols to get it right:

x R y if and only if (x = y and not(x is an element of y))

That is a description of the relation R that was plugged in.

On the topic of this thread (meta-theorems asserting independence, with proofs via model theory), it should be observed regarding a set m such that …

For all x,
(x is not an element of x) if and only if (x,x) is an element of m.

… that m could be used as a representation in a model for the predicate “x is not an element of x.”