Four labeled parts have sizes 1, 2, 3 and 6. A part can be assembled into another exactly when its size divides the other size. Here “below” means divides, not “is numerically smaller.” For example, 2 and 3 cannot be compared by divisibility even though their ordinary numerical sizes can be compared.
Draw the allowed comparisons before reading the diagram. Include a comparison of every part with itself. The diagram compresses the result: upward routes show comparisons, and a point also compares with itself without a drawn loop.
Complete numerical equivalent: 1 divides 1, 2, 3, 6; 2 divides 2, 6; 3 divides 3, 6; 6 divides 6. The route from 1 to 6 represents a comparison even though no direct line joins them.
Predict
Remove 1, keeping the same divisibility rule on the remaining elements. Predict whether there is still a bottom element. Both 2 and 3 now have nothing strictly below them, but neither lies below the other. They are minimal, and neither is least. A least element must lie below every element in the domain. A minimal element only forbids another element strictly below itself.
Similarly, maximal means nothing strictly above, while greatest means above everything. In the original four-element example, 1 is least and 6 is greatest. Remove 6 instead and both 2 and 3 become maximal, with no greatest element. Removing a point changes the domain; it does not create a new comparison between surviving points.
Experiment
Make a four-by-four grid with row and column labels 1, 2, 3, 6. Mark the entry in row a and column b when a divides b. There should be nine marked entries. Now erase the four diagonal entries. The remaining five describe strict divisibility: a divides b and a is different from b.
Test three properties directly. Every diagonal entry was present in the first grid. Whenever both a-to-b and b-to-a were present, the labels were equal. Whenever a-to-b and b-to-c were present, a-to-c was present too. These properties are reflexivity, antisymmetry and transitivity. Antisymmetry permits self-comparisons; it does not say comparisons can never run in both directions.
Change the domain to the two integers 1 and −1. Each divides the other, although they are different. Your earlier antisymmetry conclusion used the positive-integer domain. A relation’s familiar name does not prove its axioms on every domain.
Understand
A non-strict partial order ≤ is reflexive (x≤x), antisymmetric (x≤y and y≤x imply x=y), and transitive (x≤y and y≤z imply x≤z). A strict partial order ≺ is irreflexive (never x≺x) and transitive. The symbols are names for relations; they need not denote ordinary numerical order. Some sources call the non-strict relation a weak order. Here that phrase must not be confused with a strict weak order, a separate notion that also requires transitivity of incomparability.
Starting with ≺, define x≤y to mean x=y or x≺y. Reflexivity follows from equality. If x≤y and y≤x with distinct elements, then x≺y≺x would imply x≺x, a contradiction. Thus antisymmetry holds. For transitivity, an equality in either premise simply substitutes one label; if neither premise is equality, use transitivity of ≺. This proves all three axioms, rather than just observing them in our grid.
Conversely, start with ≤ and define x≺y when x≤y and x≠y. Irreflexivity follows immediately. If x≺y and y≺z, transitivity gives x≤z. If x=z, the premises give x≤y≤x, so antisymmetry forces x=y, contradicting x≺y. Therefore x≠z and x≺z. Adding or removing exactly the diagonal undoes the other construction.
In a finite nonempty poset, a unique minimal element m is least. To prove this, choose any x. If it is not minimal, move strictly downward; repeat while possible. No element can repeat, since a strict cycle would give a self-comparison by transitivity. Finiteness therefore forces the process to stop at a minimal element y≤x. Uniqueness gives y=m. Because x was arbitrary, m≤x for every x. Reversing the order proves the corresponding theorem for a unique maximal element.
This proof explains its own limitation. Take all integers in their usual order and add an isolated element a that compares only with itself. Every integer has a smaller integer, so a is the only minimal element. Yet a is not least: it does not compare with any integer. The downward process can continue forever in the integer component.
Simultaneous axioms force a specific relation
On a three-point set, equality relates each point only to itself. It is both an equivalence relation and a non-strict partial order. Is another relation possible? If R is symmetric and antisymmetric, xRy implies yRx by symmetry, then x=y by antisymmetry. Therefore R is contained in the diagonal Δ, consisting of all pairs (x, x) with x in S. Conversely every subset of the diagonal is symmetric and antisymmetric, and is automatically transitive. Adding reflexivity forces R=Δ. Thus the only relation that is both an equivalence relation and a partial order is equality. This proof works on any set, including the empty set; checking a few examples would not prove uniqueness. Without reflexivity, a proper subdiagonal relation is possible.
Can a relation be both an equivalence and an order?
Toggle a pair (row, column), then inspect an explicit failed axiom. Predict which mutual-relatedness classes survive and whether the quotient is justified.
R
a
b
c
a
b
c
reflexive: Passes
symmetric: Fails; witness (a, c)
antisymmetric: Fails; witness (a, b)
transitive: Passes
Equivalence relation: no. Partial order: no.
Preorder verified. Classes: {a, b}; {c}. Quotient comparisons include equality, not just cover edges.
Class ≤ class
a, b
c
a, b
Yes
Yes
c
No
Yes
A missing reflexivity witness names one point; symmetry/antisymmetry witnesses name two; transitivity names (x,y,z) with xRy and yRz but not xRz. The editor checks one finite relation, not the general proof. Temporary nine-bit state; no automatic mastery.
Keep indistinguishable points together: from preorder to partial order
A preorder is reflexive and transitive, but need not be antisymmetric. For example, a and b may relate in both directions while both lie below c. Define x∼y when xRy and yRx. Reflexivity of R gives x∼x; the definition is symmetric; if x∼y and y∼z, transitivity of R gives both xRz and zRx. Hence ∼ is an equivalence relation, partitioning the set into classes.
Define [x]≤[y] when xRy. First prove this is independent of representatives. If x′∼x and y′∼y, then x′Rx and yRy′; together with xRy, transitivity yields x′Ry′. Reversing the roles gives the converse, so the truth of the comparison is unchanged by either replacement. Reflexivity and transitivity now pass to classes. If [x]≤[y] and [y]≤[x], then x∼y, so their classes are equal: antisymmetry holds. The quotient is therefore a partial order.
In the worked preorder with a∼b and both below c, the classes are the class containing a and b, and the singleton class containing c, forming a two-element chain. The universal relation has one class and a one-element quotient. If the original preorder is symmetric, any comparison xRy already gives x∼y, so the quotient order is equality. If it is antisymmetric, every class is a singleton and the quotient changes only notation. These boundary cases connect the construction with the simultaneous-axiom theorem rather than treating them as unrelated definitions.
The hypotheses matter: a reflexive relation with aRb, bRa, bRc and cRb but no aRc or cRa has mutual-relatedness that is not transitive. It cannot be partitioned into the claimed equivalence classes by simply grouping mutual pairs. A quotient formula written on unproved classes is not a well-defined construction.
Apply
On the three elements p, q, r, allow only the strict comparison p≺r. Is this a strict partial order? Yes: it is irreflexive, and there is no two-step chain needing another comparison. But p is incomparable with q and q is incomparable with r, while p and r are comparable. Incomparability need not be transitive in a partial order.
Write the associated non-strict relation in full: its pairs are (p,p), (q,q), (r,r), (p,r). Identify both minimal elements and both maximal elements. The minima are p and q; the maxima are q and r. The isolated q is both minimal and maximal without being least or greatest. Explain why this is consistent with the definitions instead of treating the words as synonyms.
Continue
Carry the grid, the strict relation and the extrema explanations into the written practice. The next representation lesson will justify why upward routes can replace a long list of comparisons in a finite diagram. Keep two questions separate: which relation is being defined, and which picture or data structure represents it? A successful finite grid check is evidence about that grid; the quantified proofs above explain the general conversions and their domain assumptions.
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Classify all relations on an arbitrary set that are both symmetric and antisymmetric. Then classify those that are both equivalence relations and non-strict partial orders. Handle the empty set and explain which axiom forces the full diagonal rather than a subset.
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Let R be a reflexive transitive relation. Define x∼y iff xRy and yRx. Prove this is an equivalence relation and that [x]≤[y] iff xRy gives a well-defined partial order on its classes. Explain what the construction does when R is symmetric, and when R is antisymmetric.
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Order D={2,3,6,12,18} by divisibility: a≤b means a divides b. Prove this is a non-strict partial order. List all minimal and maximal elements, decide whether least and greatest elements exist, and exhibit incomparable pairs.
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Let < be a transitive irreflexive relation on S. Define x≤y iff x=y or x<y. Prove ≤ is a reflexive antisymmetric transitive relation. Conversely, from any reflexive antisymmetric transitive ≤, define x<y iff x≤y and x≠y; prove transitivity and irreflexivity. Show the constructions undo each other.
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A learner says antisymmetric means that whenever aRb, bRa is false. Explain the error using equality on a nonempty set. Then test divisibility as a non-strict order on all integers, compared with the nonnegative integers. Include zero in your reasoning.
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Prove that a nonempty finite partially ordered set with exactly one minimal element m has m as its least element. Explain why merely saying “nothing is smaller than m” is not a proof. State the dual result for a unique maximal element.
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Let S be the disjoint union of the integers and a new symbol a. Order integers by their usual ≤, relate a to itself, and make a incomparable with every integer. Prove this is a partial order with a unique minimal element but no least element. Explain which step of the finite theorem fails.
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On {a,b,c}, let the only strict comparison be a<c. Prove this is a strict partial order. Show that incomparability is not transitive. Explain why “non-strict (weak) partial order” in this lesson must not be confused with the separate term “strict weak order,” which adds transitive incomparability to a strict order.
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Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
