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Since set theory is proven to be inconsistent by Russell, what is the method mathematicians use to define the equality, the "greater than" and "the smaller than"relationships between two natural numbers without taking the cardinalities of sets into account?

For instance, I could state that a set containing 2 elements is smaller than a set containing 3 of them because there is no surjective function that has the latter set as its domain and the former as its image.

How can I accomplish a solid definition of these relationships without set theory?

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3 Answers 3

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If you're speaking in the language of Peano arithmetic, you can define $a < b$ recursively as:

  1. $0 < S(0)$
  2. $\forall x, y : (x < y) \to (x < S(y))$
  3. $\forall x, y : (x < y) \to (S(x) < S(y))$

where $S(x)$, the successor of $x$, can be read as "$x+1$".

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  • $\begingroup$ Thanks. Doesn't this definition shift the problem to the definition of successor or addition (+)? $\endgroup$ Dec 3, 2017 at 18:10
  • $\begingroup$ "Doesn't this definition shift the problem to the definition of successor or addition (+)?" Yes, but that wasn't what you asked. The peano postulates do not rely as cardinality or functions between sets but on axioms building up numbers from 0. Definition of addition and successor can be researched. $\endgroup$
    – fleablood
    Dec 3, 2017 at 18:17
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First, as Noah Schweber remarked in his comment, Russell only showed that a particular theory was inconsistent, not that set theory as a whole is inconsistent.

Second, we can define $x < y$ in the natural numbers as: $\exists z (z \not = 0 \wedge x+z=y)$.

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  • $\begingroup$ Thank you, doesn't this definition require a strong definition of + (addition)? $\endgroup$ Dec 3, 2017 at 18:08
  • $\begingroup$ @AntonMariaPrati What do you mean? What is a "strong" definition, and why do you think my answer needs one? $\endgroup$
    – Nagase
    Dec 3, 2017 at 18:15
  • $\begingroup$ You didn't ask for a strong definition of addition. Just a definition of "<" then that didn't rely on cardinality of sets. This one doesn't so... the question asked was answered. $\endgroup$
    – fleablood
    Dec 3, 2017 at 18:20
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"Since set theory is proven to be inconsistent by Russell,"

As Noah Schweber pointed out. One set theory was proven inconsistent. Modern set theory does not have that error.

And That one aspect of one set theory was inconsistent does not mean all are.

And, I'm not sure, but I think we didn't even define "less than" in terms of set theory until after Russell. I could be wrong.

Not as elegant as singerng's or Nagase's (which are more or less equivalent) one can do it axiamatically be field order axioms:

At order $"<"$ is a relation where: i) exactly one of $a < b; b< a; a=b$ is true ii) $a<b;b< c \implies a< c$. A field is defined by the usual axioms: addition and multiplication are commutative and associative and together are distributive; there are elements 1 and 0 that act as multiplicative and additive identities; there are multiplicative inverses. An ordered field has the following two axioms:

i) $a<b \implies a+c < a+d$

ii) $a>0; b> 0\implies ab > 0$.

With the concept of induction (basically the Peano postulates which made singerng's or Nagase's answers possible), we have gone beyond the natural numbers to all the rationals.

Now you may argue that a Field is, by definition, a set and that the natural numbers are a set but the only aspect of set theory used is simply that we can consider items as a group. That was not the contradiction. And if we cant consider items as a group, mathematics is pretty pointless. It's a bit like saying since individual opinion is fallable is it possible to do philosophy without thinking. Um....

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  • $\begingroup$ I, in fact, was assuming that sets are the same objects as groups, given each element of a group is distinct. Thanks for the thorough explanation. $\endgroup$ Dec 3, 2017 at 19:00
  • $\begingroup$ We shouldn't say "sets are the same object as groups" as "group" has a specific meaning. My point is, that we can group things together into collections isn't "set theory". "set theory" is analysis on the abstract concept of the collections. If we can't group things together we have nothing. Not even language. $\endgroup$
    – fleablood
    Dec 3, 2017 at 21:50

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