Skip to main content
Clarified how answer relates to negating axioms
Source Link
Joe Lamond
  • 1.7k
  • 1
  • 12
  • 20

I just wanted to point out as a comment that there is some ambiguity, already mentioned by Joel David Hamkins, about what it means to "negate fundamental axioms" (there is some further discussion here). The issue is not with the negation per se, but rather with what it means to "subtract" axioms from a theory.

Let me try to give a "toy example"toy example of what can go wrong. A strict partial order $<$ on a set $X$ may be defined as a relation on $X$ satisfying the following three axioms:

  1. Irreflexivity: there does not exist an $a\in X$ such that $a<a$.
  2. Asymmetry: for all $a,b\in X$, if $a<b$ then it is not the case that $b<a$.
  3. Transitivity: for all $a,b,c\in X$, if $a<b$ and $b<c$, then $a<c$.

It turns out, however, that the asymmetry axiom is redundant, in the sense that if a relation $<$ on $X$ is irreflexive and transitive, then $<$ is asymmetric. Thus, we could have defined a strict partial order to simply be a irreflexive and transitive relation.

Now, if someone speaks about "the theory $T$ of strict partial orders without the irreflexive axiom", then what could they mean? If we axiomatise strict partial orders as being relations satisfying (1)-(3), then $T$ is identical to the theory of strict partial orders, since irreflexivity can be deduced from asymmetry (take $a=b$ in (2)). On the other hand, if we axiomatise strict partial orders as being irreflexive and transitive relations, then $T$ is simply the theory of transitive relations, which is clearly not the same as the theory of strict partial orders.

If we go further and consider "the theory $S$ of strict partial orders with the irreflexive axiom replaced by its negation", then either $S$ is an inconsistent theory, or $S$ is the theory of transitive relations $<$ for which there exists an $a$ such that $a<a$, depending on what is exactly meant by $S$.

Similar problems arise in set theory. In a number of set theory texts, the axiom of pairing is omitted from the $\mathsf{ZFC}$ axioms, since it can be deduced from replacement. But this presents an issue when we want to consider Zermelo's theory $\mathsf Z$, since we want $\mathsf Z$ to have pairing, but not replacement. Describing $\mathsf Z$ as "$\mathsf{ZFC}$ with replacement and regularity removed" can easily lead to confusion.

I just wanted to point out as a comment that there is some ambiguity, already mentioned by Joel David Hamkins, about what it means to "negate fundamental axioms" (there is some further discussion here).

Let me try to give a "toy example" of what can go wrong. A strict partial order $<$ on a set $X$ may be defined as a relation on $X$ satisfying the following three axioms:

  1. Irreflexivity: there does not exist an $a\in X$ such that $a<a$.
  2. Asymmetry: for all $a,b\in X$, if $a<b$ then it is not the case that $b<a$.
  3. Transitivity: for all $a,b,c\in X$, if $a<b$ and $b<c$, then $a<c$.

It turns out, however, that the asymmetry axiom is redundant, in the sense that if a relation $<$ on $X$ is irreflexive and transitive, then $<$ is asymmetric. Thus, we could have defined a strict partial order to simply be a irreflexive and transitive relation.

Now, if someone speaks about "the theory $T$ of strict partial orders without the irreflexive axiom", then what could they mean? If we axiomatise strict partial orders as being relations satisfying (1)-(3), then $T$ is identical to the theory of strict partial orders, since irreflexivity can be deduced from asymmetry (take $a=b$ in (2)). On the other hand, if we axiomatise strict partial orders as being irreflexive and transitive relations, then $T$ is simply the theory of transitive relations, which is clearly not the same as the theory of strict partial orders.

Similar problems arise in set theory. In a number of set theory texts, the axiom of pairing is omitted from the $\mathsf{ZFC}$ axioms, since it can be deduced from replacement. But this presents an issue when we want to consider Zermelo's theory $\mathsf Z$, since we want $\mathsf Z$ to have pairing, but not replacement. Describing $\mathsf Z$ as "$\mathsf{ZFC}$ with replacement and regularity removed" can easily lead to confusion.

I just wanted to point out as a comment that there is some ambiguity, already mentioned by Joel David Hamkins, about what it means to "negate fundamental axioms" (there is some further discussion here). The issue is not with the negation per se, but rather with what it means to "subtract" axioms from a theory.

Let me try to give a toy example of what can go wrong. A strict partial order $<$ on a set $X$ may be defined as a relation on $X$ satisfying the following three axioms:

  1. Irreflexivity: there does not exist an $a\in X$ such that $a<a$.
  2. Asymmetry: for all $a,b\in X$, if $a<b$ then it is not the case that $b<a$.
  3. Transitivity: for all $a,b,c\in X$, if $a<b$ and $b<c$, then $a<c$.

It turns out, however, that the asymmetry axiom is redundant, in the sense that if a relation $<$ on $X$ is irreflexive and transitive, then $<$ is asymmetric. Thus, we could have defined a strict partial order to simply be a irreflexive and transitive relation.

Now, if someone speaks about "the theory $T$ of strict partial orders without the irreflexive axiom", then what could they mean? If we axiomatise strict partial orders as being relations satisfying (1)-(3), then $T$ is identical to the theory of strict partial orders, since irreflexivity can be deduced from asymmetry (take $a=b$ in (2)). On the other hand, if we axiomatise strict partial orders as being irreflexive and transitive relations, then $T$ is simply the theory of transitive relations, which is clearly not the same as the theory of strict partial orders.

If we go further and consider "the theory $S$ of strict partial orders with the irreflexive axiom replaced by its negation", then either $S$ is an inconsistent theory, or $S$ is the theory of transitive relations $<$ for which there exists an $a$ such that $a<a$, depending on what is exactly meant by $S$.

Similar problems arise in set theory. In a number of set theory texts, the axiom of pairing is omitted from the $\mathsf{ZFC}$ axioms, since it can be deduced from replacement. But this presents an issue when we want to consider Zermelo's theory $\mathsf Z$, since we want $\mathsf Z$ to have pairing, but not replacement. Describing $\mathsf Z$ as "$\mathsf{ZFC}$ with replacement and regularity removed" can easily lead to confusion.

Source Link
Joe Lamond
  • 1.7k
  • 1
  • 12
  • 20

I just wanted to point out as a comment that there is some ambiguity, already mentioned by Joel David Hamkins, about what it means to "negate fundamental axioms" (there is some further discussion here).

Let me try to give a "toy example" of what can go wrong. A strict partial order $<$ on a set $X$ may be defined as a relation on $X$ satisfying the following three axioms:

  1. Irreflexivity: there does not exist an $a\in X$ such that $a<a$.
  2. Asymmetry: for all $a,b\in X$, if $a<b$ then it is not the case that $b<a$.
  3. Transitivity: for all $a,b,c\in X$, if $a<b$ and $b<c$, then $a<c$.

It turns out, however, that the asymmetry axiom is redundant, in the sense that if a relation $<$ on $X$ is irreflexive and transitive, then $<$ is asymmetric. Thus, we could have defined a strict partial order to simply be a irreflexive and transitive relation.

Now, if someone speaks about "the theory $T$ of strict partial orders without the irreflexive axiom", then what could they mean? If we axiomatise strict partial orders as being relations satisfying (1)-(3), then $T$ is identical to the theory of strict partial orders, since irreflexivity can be deduced from asymmetry (take $a=b$ in (2)). On the other hand, if we axiomatise strict partial orders as being irreflexive and transitive relations, then $T$ is simply the theory of transitive relations, which is clearly not the same as the theory of strict partial orders.

Similar problems arise in set theory. In a number of set theory texts, the axiom of pairing is omitted from the $\mathsf{ZFC}$ axioms, since it can be deduced from replacement. But this presents an issue when we want to consider Zermelo's theory $\mathsf Z$, since we want $\mathsf Z$ to have pairing, but not replacement. Describing $\mathsf Z$ as "$\mathsf{ZFC}$ with replacement and regularity removed" can easily lead to confusion.