I found this proof that points out the difficulty of proving the 3rd axiom(Ongley & Carey,Russell-A Guide for the perplexed): Summarized:in a finite universe,say with 10 things,the successor of 10 is null,because there is no class of 11 things and the successor of 11 is likewise null.Then 10+1=11+1.Peano axiom 3 holds for an infinite universe.

I'm having a difficulty on accepting the proof since the notion of null is vague in my opinion,hence a difficulty also on 10+1=11+1. Unless this can be explained to be a valid proof is there a better way to prove that Peano's 3rd axiom is inconsistent in a finite universe?

  • $\begingroup$ I don't like the bit about the successor of 10 being null. From Peano's Axioms, we have a function mapping $N$ to itself that is both injective but not surjective. Therefore, $N$ is Dedekind-infinite by definition. $\endgroup$ – Dan Christensen Nov 8 '16 at 15:32

See : - Bertrand Russell, Introduction to Mathematical Philosophy (1919).

[page 18] The number of a class is the class of all those classes that are similar to it.

[page 23] $0$ is the class whose only member is the null-class [the empty set].

The successor of the number of terms in the class $\alpha$ is the number of terms in the class consisting of $\alpha$ together with $x$, where $x$ is any term not belonging to the class.

In set terms, if $10=\text {No}(\alpha)$, where $\alpha$ has $10$ elements, then $11=\text {No}(\alpha \cup \{ x \})$ where : $x \notin \alpha$.

If the total number of objects in the universe is $10$ we have no class with $11$ objects and thus the class of all classes with $11$ elements will be empty.

If so :

$11 = \text {No}(\emptyset)= 12 = \ldots ...$

violating the axiom : “no two numbers have the same successor.”

[page 24] Thus we should have $11 = 12$; therefore the successor of $10$ would be the same as the successor of $11$, although $10$ would not be the same as $11$. Thus we should have two different numbers with the same successor. This failure of the third axiom cannot arise, however, if the number of individuals in the world is not finite.

Note that Russell refers to the original Peano's formulation (1889) : skipping the equality axioms (from 2 to 5) axiom 7 reads :

if $a, b \in \mathbb N$, then : $a=b$, iff $a+1=b+1$

that can be read : two numbers are equal if and only if they have the same successors.


Given a set $N$ and a function $s\colon N \to N$, Peano's axioms say (among other things):

  1. The function $s$ is injective, i.e. if $s(x) = s(y)$ then $x=y$
  2. There exists an element $o \in N$ such that for all $x \in N$ we have $f(x) \neq o$. In other words, the function $f$ is not surjective.

These two things cannot hold simultaneously if $N$ is a finite set. Indeed, we have the following

Theorem. Let $S$ be a finite set, let $f\colon S\to S$ be an injective function. Then $f$ is surjective.

Proof. We use induction on the size $n$ of $S$. If $n=1$, then every function $S \to S$ is surjective. So there is nothing to prove. Now let $n$ be any number, and assume we have proven the statement for sets of size smaller than $n$. Take any element $s \in S$, and assume there do not exist $x \in S$ such that $f(x)=s$. In this case, $f$ restricts to a function $f'\colon S\setminus \{s\} \to S\setminus \{s\}$, which is automatically injective. By our induction assumption, $f'$ is surjective. In particular, there exists $x \neq s$ such that $f'(x) = f(s)$. But $f'(x)=f(x)$, and since $x \neq s$, this is a contradiction to the fact that $f$ is injective.


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