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$(\mathbb{N}, \le)$ has no maximal element: Zorn's chain hypothesis fails

Statement refuted

Refuted claim: the chain hypothesis of Zorn's lemma is redundant, that is every nonempty poset has a maximal element (Zorn's lemma, Maximal element and greatest element).

The witness is $(\mathbb{N}, \le)$ with its usual order (Order on the natural numbers). It is nonempty and has no maximal element whatsoever, and the hypothesis of Zorn's lemma that it violates is exactly one: $\mathbb{N}$ is itself a chain (Chain in a poset) and it has no upper bound in $\mathbb{N}$ (Upper bound, least upper bound, and strict upper bound).

Facts & Assumptions

Given: $\mathbb{N}$ with the order $m \le n \iff \exists k \in \mathbb{N}\ (m + k = n)$ (Order on the natural numbers) and addition satisfying $m + 0 = m$ and $m + \sigma(n) = \sigma(m + n)$ (Addition of natural numbers); $m < n$ abbreviates $m \le n$ together with $m \ne n$ .

[L1]

$\le$ is a linear order on $\mathbb{N}$ : reflexive, antisymmetric, transitive and total ( $\le$ is a linear order on $\mathbb{N}$ ).

[L2]

$n \ne \sigma(n)$ for every $n \in \mathbb{N}$ (No natural number equals its own successor).

[L3]

$m$ is maximal in a poset when no $x$ satisfies $m < x$ (Maximal element and greatest element).

[L4]

$u$ is an upper bound of $S$ when $s \le u$ for every $s \in S$ (Upper bound, least upper bound, and strict upper bound).

[L5]

A subset is a chain when any two of its elements are comparable (Chain in a poset).

[L6]

Zorn's lemma, stated under the Axiom of Choice (The Axiom of Choice), which it assumes outright: a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma). The Axiom of Choice is a standing assumption of the theorem, not a hypothesis on the poset.

Counterexample

technique · direct

1.1

$(\mathbb{N}, \le)$ is a poset and it is nonempty, since $0 \in \mathbb{N}$ .

givenL1

1.2

For every $n \in \mathbb{N}$ one has $n + \sigma(0) = \sigma(n + 0) = \sigma(n)$ , so $n \le \sigma(n)$ ; and $n \ne \sigma(n)$ , so $n < \sigma(n)$ .

givenL2

2.1

$\mathbb{N}$ has no maximal element: for any $n$ the element $\sigma(n)$ lies strictly above it.

step 1.2L3

2.2

$\mathbb{N}$ is a chain of the poset $(\mathbb{N}, \le)$ , because $\le$ is total, so any two natural numbers are comparable.

step 1.1L1L5

3.1

That chain has no upper bound in $\mathbb{N}$ : an upper bound $u$ would satisfy $n \le u$ for every $n$ , in particular $\sigma(u) \le u$ ; combined with $u \le \sigma(u)$ , antisymmetry would give $u = \sigma(u)$ , which [L2] forbids. So no $u \in \mathbb{N}$ is an upper bound of $\mathbb{N}$ .

step 2.2step 1.2L1L2L4

4.1

So $(\mathbb{N}, \le)$ is a nonempty poset with no maximal element, and of the hypotheses [L6] places on the poset the single one that fails is that every chain has an upper bound, the offending chain being $\mathbb{N}$ itself; the Axiom of Choice, assumed throughout [L6], is not a property of $(\mathbb{N}, \le)$ and is neither used nor contradicted here. The claim is refuted and Zorn's lemma is untouched.

step 1.1step 2.1step 3.1L6∎

Remarks

Nothing exotic is at work. The poset is totally ordered, it is the most familiar order there is, and it is even well ordered (The well-ordering principle). What it lacks is a ceiling. So the hypothesis Zorn's lemma really needs is boundedness of chains, and no amount of good behaviour elsewhere substitutes for it.
It fails only at the top. Every chain of $\mathbb{N}$ that has an upper bound at all has a least one: the set of its upper bounds is a nonempty subset of $\mathbb{N}$ , so The well-ordering principle hands back its least element. The empty chain has least upper bound $0$ : it is vacuously an upper bound, and $0\le u$ for every natural $u$ because $0+u=u$ (Left identity for addition, Order on the natural numbers). So the only chains without suprema are the ones with no upper bound whatever, and $\mathbb{N}$ is one of them. The same observation, read as a statement about suprema rather than upper bounds, is A progressive map with no fixed point, on a poset that is not chain-complete.
Nonemptiness is not what fails here, and how much work it does depends on the convention. Under the convention of Chain in a poset, where $\emptyset$ counts as a chain, "every chain has an upper bound" already forces $P \ne \emptyset$ , since an upper bound of $\emptyset$ is just some element of $P$ ; the separate nonemptiness hypothesis of Zorn's lemma is then emphasis rather than extra strength. Under the competing convention, where "chain" means nonempty chain, the empty poset satisfies the chain hypothesis vacuously and has no maximal element, so nonemptiness must be assumed outright. Either way, what $(\mathbb{N}, \le)$ isolates is the failure of the chain hypothesis alone.
Adding a single element $\infty$ above every natural number repairs everything: the chain $\mathbb{N}$ then has upper bound $\infty$ , every chain has one, and $\infty$ is the maximal element Zorn's lemma promises.

Depends on

Used by

A progressive map with no fixed point, on a poset that is not chain-complete Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

I. Khatchatourian, The Axiom of Choice (University of Toronto MAT327 notes) (standard reference, not scraped)
Encyclopedia of Mathematics, Zorn lemma (standard reference, not scraped)
Zorn's lemma (Wikipedia) (standard reference, not scraped)
Upper and lower bounds (Wikipedia) (standard reference, not scraped)

(N,≤)(\mathbb{N}, \le)(N,≤) has no maximal element: Zorn's chain hypothesis fails