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False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)
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FALSE: successor-closure alone forces a set to be all of N\mathbb{N}

Statement

False statement. If a set S⊆NS \subseteq \mathbb{N} is nonempty and closed under the successor (n∈S⇒σ(n)∈Sn \in S \Rightarrow \sigma(n) \in S), then S=NS = \mathbb{N}. (That is, the induction principle would hold without its base case 0∈S0 \in S.)

Facts & Assumptions

Given: the claim above.

[L1]

The induction principle requires 0∈S0 \in S (The principle of mathematical induction).

[L2]

σ(n)≠0\sigma(n) \neq 0 for all nn (P1) (The von Neumann naturals form a Peano system).

Refutation

technique · direct
1.1

Take S=N∖{0}={n∈N:n≠0}S = \mathbb{N} \setminus \{0\} = \{n \in \mathbb{N} : n \neq 0\}, the set of nonzero naturals; it is nonempty (for instance 1=σ(0)∈S1 = \sigma(0) \in S).

given
1.2

SS is closed under σ\sigma: for any nn, σ(n)≠0\sigma(n) \neq 0 by P1 [L2], so σ(n)∈S\sigma(n) \in S.

L2
2.1

But 0∉S0 \notin S, so S≠NS \neq \mathbb{N}; the nonempty successor-closed set SS is not all of N\mathbb{N}, refuting the claim.

step 1.1step 1.2
3.1

The base case 0∈S0 \in S is therefore indispensable in the induction principle [L1]; successor-closure and nonemptiness do not suffice.

step 2.1L1∎

Depends on

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Sources