Symmetric and Transitive Implies Reflexive- The Mathematical Proof

What the Theorem Actually Says

There's a classic result in relation theory that trips up students constantly:

If a relation is symmetric and transitive, it must also be reflexive.

Most textbooks present this as a three-line proof. Most students walk away thinking they understand it. They're usually wrong.

The theorem is technically correct. But it hides a crucial assumption that makes the whole thing almost useless without context.

The Setup

Let R be a relation on a set A. R is symmetric when:

∀a,b ∈ A: if aRb then bRa

R is transitive when:

∀a,b,c ∈ A: if aRb and bRc then aRc

R is reflexive when:

∀a ∈ A: aRa

Simple enough. Now here's the proof everyone learns.

The Proof

Take any element a ∈ A.

Since R is symmetric and transitive, we have:

• From symmetry: aRb → bRa
• From transitivity: aRb and bRa → aRa

Therefore aRa for all a ∈ A.

QED.

Except this proof has a massive hole.

Where the Proof Breaks Down

The argument assumes that for every a ∈ A, there exists some b ∈ A with aRb.

But nothing in symmetry or transitivity guarantees this.

Consider the counterexample that destroys the naive version:

Counterexample

Let A = {1, 2, 3}

Let R = {(1, 1), (1, 2), (2, 1), (2, 2)}

Check the properties:

• Symmetric? Yes. 1R2 and 2R1 both exist. 1R1 and 2R2 are self-pairs.
• Transitive? Yes. 1R2 and 2R1 gives 1R1. 2R1 and 1R2 gives 2R2. All other combinations work.
• Reflexive? No. 3R3 does not exist.

Element 3 sits in A but has no relations at all. The proof assumes every element participates in at least one relation. This assumption is not implied by symmetry or transitivity.

The Correct Theorem

The proper statement includes the missing condition:

Theorem: Let R be a relation on a non-empty set A. If R is symmetric and transitive, and every element of A is related to at least one element (i.e., ∀a ∈ A, ∃b ∈ A: aRb), then R is reflexive.

This condition—every element participates in at least one ordered pair—is sometimes called seriality or left-total. It's not automatically true for symmetric transitive relations.

Why This Matters

Understanding this distinction matters for two reasons:

• Proof writing: If you state the theorem without the seriality condition, your proof is incomplete. Any instructor will deduct points.
• Understanding equivalence relations: An equivalence relation requires reflexivity, symmetry, and transitivity. The theorem shows that symmetry + transitivity + participation = reflexivity. But you still need the participation condition explicitly.

Common Misconceptions

Students often believe one of these false statements:

• "Symmetric and transitive always imply reflexive" — False without the participation condition
• "The proof is trivial and always works" — False; the hole in the proof is the existence of related elements
• "This theorem makes reflexivity redundant" — False; reflexivity is still a separate condition for full equivalence relations

When the Theorem Does Apply

The theorem becomes useful in specific contexts where participation is guaranteed:

• Graphs: If you have a connected component, every vertex in that component has at least one edge
• Function composition: When dealing with functional relations where the domain equals the codomain
• Partial orders with coverage: In certain lattice structures where every element is covered by something

Summary Table

Getting Started: How to Apply This

When you're given a relation and asked to analyze it:

1. Check participation first. List all elements. For each one, check if it appears in at least one ordered pair. If any element is isolated, the theorem doesn't apply.
2. Verify symmetry. For every pair (a,b), confirm (b,a) exists.
3. Verify transitivity. For every chain aRb and bRc, confirm aRc exists.
4. Apply the theorem only if participation holds. Then you can conclude reflexivity.

Example walkthrough:

A = {1, 2, 3}

R = {(1,1), (1,2), (2,1), (2,2), (3,3)}

• Participation: All elements appear. ✓
• Symmetry: 1R2 implies 2R1. ✓
• Transitivity: 1R2 and 2R1 gives 1R1. ✓
• Conclusion: R is reflexive. ✓

The Bottom Line

The theorem "symmetric + transitive implies reflexive" is true. But it's not the free gift it appears to be. The participation condition is the real work, and it's not automatic.

When you see this theorem in a textbook or exam, your first thought should be: "Where's the condition that every element participates?" That's the catch. That's what makes the proof work.

Once you see it, the theorem becomes trivial. Before you see it, you might write a broken proof and wonder why you lost points.