8

The standard way of using ZFC to encode the rest of mathematics is sometimes criticized because it introduces unnecessary, strange properties such as, for example $1\in 2$ if we encode integers by ordinals.

From a constructivist type theorist's perspective, uples or functions or complex numbers should have nothing to do with the membership relation, we should have some sort of "free algebra".

Formally, consider a set $M$, a binary relation $\varepsilon$ on $M$, and a partition $S\cup P \cup C \cup F$ of $M$. The intention is that $S$ represents sets, $P$ represents pairs, $C$ represents complex numbers and $F$ represents functions.

I say that $M$ is a "free model" of usual mathematics if

(1) $(M,\varepsilon)$ satisfies all the axioms of ZFC except extensionality.

(2) $(S,\varepsilon)$ satisfies all the axioms of ZFC. In partcular, extensionality is satisfied on $S$ : If two elements of $S$ have the same $\varepsilon$-elements (in $M$), they are equal.

(3) All the elements in $P,C,F$ are $\varepsilon$-empty.

(4) There is a bijection $M^2 \to P$.

(5) There is a bijection ${\mathbb C} \to C$.

(6) There is a bijection ${\cal F}(M) \to F$, where ${\cal F}(M)$ is the set of all partial functions from $M$ to $M$, i.e. the set of all $A\subseteq M^2$ satisfying $(a,b)\in A,(a,b')\in A \Rightarrow b=b'$.

Note that the bijection in (6) is necessarily outside $M$.

Let us assume that ZFC is consistent. Does it necessarily follow that there is a "free model" of usual mathematics ?

I thought of starting with a countable model and using transfinite induction, but it is not clear how one should define pairs or functions in the new model.

Ewan Delanoy
  • 61,600
  • If this question receives no answers, consider rephrasing it to be more precise and add it to Stack Overflow – theREALyumdub Sep 12 '15 at 06:09
  • 1
    @theREALyumdub what's not precise enough for you ? And you probably mean MathOverflow, not StackOverflow – Ewan Delanoy Sep 12 '15 at 07:05
  • 1
    Yes that is what I mean lol – theREALyumdub Sep 12 '15 at 13:07
  • My suggestion was to make it more clear what was intended, because the subject material is not exactly accessible to most people as far as knowledge goes. Not a lot of people can answer this question on this site so perhaps be more clear for less precise mathematicians to understand, like giving the shortest explanation of ZFC https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory – theREALyumdub Sep 12 '15 at 13:10
  • "Unnecessary, strange properties such as, for example $1\in 2$ if we encode integers by ordinals" ? Such properties are not typical for ZFC. If you take a look at this answer , then you will notice that - even in the most practical set theory on earth - there are memberships like $1\in 2$ . So I'd rather think that such "strange properties" are inevitable. – Han de Bruijn Sep 12 '15 at 15:36
  • 1
    @HandeBruijn Which is why some people abandon set theories entirely. – Ptharien's Flame Sep 12 '15 at 17:14
  • @Ptharien'sFlame: I wouldn't go that far. But my favorite set theory (see reference) is the one where sets and natural numbers are on equal footing - i.e. no set theory as a foundation of mathematics - such as with bit patterns in a computer. – Han de Bruijn Sep 12 '15 at 18:32

0 Answers0