Here is a "proof" that looks airtight on first reading. Let R be symmetric and transitive on a set A. Pick any a \in A. Pick any b with (a, b) \in R. Symmetry gives (b, a) \in R. Transitivity applied to the chain a \to b \to a gives (a, a) \in R. So every element of A is related to itself. R is reflexive. QED.
This argument is wrong, and the error is so subtle that generations of students have lost marks to it in JEE and board exams. Finding the hole is a rite of passage.
The bug: the hidden assumption
Look at the first move: "pick any b with (a, b) \in R."
What if no such b exists?
If a is an element of A that is not related to anything (not to itself, not to any other element), then the chain a \to b \to a cannot even start. There is no b to pick. The whole argument evaporates, and you never conclude (a, a) \in R.
The "proof" silently assumes every element has something attached to it. When that assumption fails, reflexivity can fail too.
Why: an implication "if (a, b) \in R and (b, a) \in R then (a, a) \in R" is vacuously satisfied when the hypothesis is empty. It gives you no information about (a, a) when a has no partners. The argument only produces self-loops at elements that already participate somewhere.
Counter-example 1: the empty relation
Let A = \{1, 2\} and R = \varnothing — the relation with no pairs at all.
Symmetric? Yes — vacuously. The defining implication "(a, b) \in R \implies (b, a) \in R" has no pair (a, b) to test, so it is automatically satisfied.
Transitive? Yes — vacuously, for the same reason. No chains of length two exist, so there is nothing to check.
Reflexive? No. For R to be reflexive on \{1, 2\}, both (1, 1) and (2, 2) must be in R. Neither is.
So R = \varnothing is symmetric and transitive but not reflexive. The "proof" at the top of this page is broken.
Counter-example 2: a partial relation
The empty relation feels like cheating. Here is a non-empty counter-example.
Let A = \{1, 2, 3\} and R = \{(1, 1), (1, 2), (2, 1), (2, 2)\}.
Check the three properties.
Symmetric. Every pair and its reverse are both in: (1,1) and (1,1); (1,2) and (2,1); (2,2) and (2,2). Yes.
Transitive. Any chain (a,b), (b,c) \in R — all of a, b, c must be in \{1, 2\}. Check each:
- (1,1) and (1,2) → need (1, 2): in.
- (1,2) and (2,1) → need (1, 1): in.
- (2,1) and (1,2) → need (2, 2): in.
- All other chains are equally fine.
Yes, transitive.
Reflexive? We need (1,1), (2,2), and (3,3) all in R. The first two are in. The third, (3,3), is not. So R is not reflexive.
This R is symmetric and transitive but not reflexive — and it is not even the empty relation.
What goes wrong in the "proof" for a = 3?
Apply the "proof" step by step to the element a = 3.
"Pick any b with (3, b) \in R."
But look at the list of pairs: R = \{(1,1), (1,2), (2,1), (2,2)\}. The first coordinate 3 appears nowhere. There is no b with (3, b) \in R. The proof asks you to pick something that does not exist, and falls silent.
The conclusion (3, 3) \in R is never produced, and in fact it is false. So the "proof" does not actually prove what it claimed.
The correct statement
A correct statement requires one extra condition:
When symmetric + transitive *does* imply reflexive
If R is symmetric, transitive, and every element of A appears somewhere in R (that is, for every a \in A there is at least one b with (a, b) \in R), then R is reflexive on A.
That extra assumption is sometimes called the seriality of R or "R has full domain." With it, the original chain-argument really does work: pick b with (a, b) \in R, use symmetry to get (b, a), then transitivity to get (a, a).
Without it, the argument cannot start.
How to check the three properties on an exam
The temptation to skip reflexivity when you have already verified symmetric and transitive is a pure trap. Reflexivity is a separate property with its own check: for every a \in A, is (a, a) \in R? You must look at every element of A and inspect whether the diagonal pair is in R.
Three independent questions. Three independent answers. A JEE question that builds on this trap will almost always construct a relation where reflexivity is what you have to catch.
One-line take-away
"Symmetric + transitive \Rightarrow reflexive" is false as a general rule. The empty relation, and any relation that leaves some element of A completely out of the picture, breaks it. Always verify reflexivity by checking the diagonal directly.
Related: Relations · Equivalence Relations · Symmetry Check Visualisation · Sets — Introduction