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A partition into intervals with non-monotone endpoints need not be a lattice congruence

Statement refuted

Statement refuted: every partition of a finite lattice into intervals (each class of the form $[d (x), u (x)]$ with endpoints in the class) is a lattice congruence.

Counterexample. In the diamond $D = {0 < a, b < 1}$ take the partition into the intervals ${0, a} = [0, a]$ , ${b} = [b, b]$ and ${1} = [1, 1]$ . It is a partition into intervals, but it is not a lattice congruence: $0 \equiv a$ and $b \equiv b$ , while $0 \lor b = b$ and $a \lor b = 1$ , so $b \equiv̸ 1$ . In terms of the criterion (The interval criterion for a lattice congruence: interval classes with monotone endpoints) the upper endpoint map fails to be order-preserving: $0 \leq b$ , but $u (0) = a \leq̸ b = u (b)$ . Thus the monotonicity hypothesis cannot be dropped, and the four-congruence count of the companion example on a chain and a diamond is a genuine restriction.

Facts & Assumptions

Given: The diamond $D = {0 < a, b < 1}$ in which $a$ and $b$ are incomparable, identified with the Boolean lattice $B ({a, b}) = {\emptyset, {a}, {b}, {a, b}}$ through $0 = \emptyset$ , $a = {a}$ , $b = {b}$ , $1 = {a, b}$ , with meet and join intersection and union (The Boolean lattice of subsets of a finite set and its rank levels); and the partition of $D$ into the blocks ${0, a}$ , ${b}$ , ${1}$ , with $[d, u] = {z : d \leq z \leq u}$ (Intervals in a poset; locally finite, lower-finite and upper-finite posets).

[F1]

In $D$ the order is inclusion and meet and join are intersection and union (The Boolean lattice of subsets of a finite set and its rank levels); hence $0 \lor b = b$ , $a \lor b = 1$ , $0 \land a = 0$ and $a \land b = 0$ .

[F2]

A partition of a set $A$ is a family of nonempty pairwise disjoint blocks whose union is $A$ , and the relation that holds between $a$ and $b$ when one block contains both is an equivalence relation whose classes are the blocks (The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

[F3]

A lattice congruence on a finite lattice is an equivalence relation with $x \equiv x^{'}$ and $y \equiv y^{'}$ implying $x \land y \equiv x^{'} \land y^{'}$ and $x \lor y \equiv x^{'} \lor y^{'}$ (Finite lattice congruences, interval endpoints and descending rooted-chain labels).

[F4]

Let $θ$ be an equivalence relation on a finite lattice whose classes are intervals $[d (x), u (x)]$ with endpoints in the class. Then $θ$ is a lattice congruence if and only if the endpoint maps $d$ and $u$ are order-preserving (The interval criterion for a lattice congruence: interval classes with monotone endpoints).

Proof

technique · direct

1.1F1F2

The three blocks are intervals with their endpoints in the block: ${0, a} = {z : 0 \leq z \leq a} = [0, a]$ , ${b} = [b, b]$ and ${1} = [1, 1]$ ; they are nonempty, pairwise disjoint, and their union is ${0, a, b, 1} = D$ . By [F2] they form a set partition of $D$ , and calling its blocks classes gives the equivalence relation $θ$ with $0 \equiv a$ , $b \equiv b$ , $1 \equiv 1$ and $0 \equiv̸ b$ , $0 \equiv̸ 1$ , $a \equiv̸ b$ , $a \equiv̸ 1$ , $b \equiv̸ 1$ .

2.1F1F3step 1.1

The relation $θ$ is not a lattice congruence: $0 \equiv a$ and $b \equiv b$ hold, but by [F1] one has $0 \lor b = b$ and $a \lor b = 1$ , and $b \equiv̸ 1$ because $b$ and $1$ lie in the distinct blocks ${b}$ and ${1}$ ; so the congruentiality requirement of [F3] for joins fails, and $θ$ is not a lattice congruence.

2.2F1step 1.1

The upper endpoint map of the partition is not order-preserving: $u (0) = a$ and $u (b) = b$ by step 1.1, and $0 \leq b$ in $D$ while $a \leq̸ b$ because $a$ and $b$ are incomparable; hence $u (0) \leq̸ u (b)$ .

3.1F4step 1.1step 2.1step 2.2∎

Conclusion. The partition of $D$ into ${0, a}$ , ${b}$ , ${1}$ is a partition into intervals with endpoints in the class (step 1.1) and its upper endpoint map is not order-preserving (step 2.2), so by the criterion [F4] it is not a lattice congruence, in agreement with the direct failure of step 2.1; this refutes the displayed statement and shows that the monotonicity hypothesis of the criterion cannot be dropped.

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Sources

Nathan Reading, Lattice congruences of the weak order: algebra, combinatorics, and geometry, Triangle Lectures in Combinatorics (2019), slides on the order-theoretic characterization of a lattice congruence (standard reference, not scraped)