How to Check Reflexivity on an Infinite Set (You Don't Check Every Element)

The definition reads: "R is reflexive if for every a \in A, (a, a) \in R."

If A is finite, you can in principle check every element. But most sets you meet in class 11 and JEE — \mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R} — are infinite. You cannot list all reals and tick them off one by one. So how do you prove that a relation on \mathbb{R} is reflexive?

You prove it the way mathematicians prove any "for every" statement on an infinite set: with a universal argument. Pick an arbitrary element, prove the property for that one element using only the general rule, and that argument automatically covers every element.

The method in one line

To check reflexivity: let a be an arbitrary element of A, and show that (a, a) satisfies the defining rule of R.

That is the whole technique. The symbol a stands for "any element" — you are not checking any specific number, you are checking the pattern. If your argument works without knowing what a is, it automatically works for every a.

Why: the statement "for every a \in A, P(a)" in mathematics is proved by taking a generic a and showing P(a) without assuming any special property of a. The argument is a template that, once written, fits every value of a simultaneously. This is standard mathematical induction-free universal proof, and it is exactly how reflexivity is proved.

Worked template

Show that "$a \le b$" on $\mathbb{R}$ is reflexive

Claim. For every a \in \mathbb{R}, (a, a) \in R (that is, a \le a).

Proof. Let a \in \mathbb{R} be arbitrary. By the definition of \le, we need a \le a. This is the statement "a is less than or equal to itself," which is true because a = a, and equality is one of the two conditions inside \le. Hence (a, a) \in R.

Since a was arbitrary, reflexivity holds for every real number.

Done. The proof never picked a specific real — it worked with the symbol a throughout, so it covers every element of \mathbb{R} in one stroke.

Notice the structure:

Start: "Let a be an arbitrary element of A."
Body: A short argument using only the rule that defines R, applied to the pair (a, a).
End: "Therefore (a, a) \in R, and since a was arbitrary, R is reflexive."

No list, no enumeration, no checking. One argument, and the universal quantifier does the rest.

Three more examples

Congruence modulo $n$ on $\mathbb{Z}$

R = \{(a, b) \mid n \mid (a - b)\} for a fixed positive integer n.

Reflexivity check. Let a \in \mathbb{Z} be arbitrary. Then a - a = 0, and 0 is a multiple of every positive integer (since 0 = n \cdot 0). So n \mid (a - a), which means (a, a) \in R.

Because a was arbitrary, R is reflexive on all of \mathbb{Z}.

"$a$ and $b$ have the same sign" on $\mathbb{R} \setminus \{0\}$

R = \{(a, b) \mid a \text{ and } b \text{ have the same sign}\}.

Reflexivity check. Let a \in \mathbb{R} \setminus \{0\} be arbitrary. Then a has the same sign as a (both are positive or both are negative — whichever a itself is). So (a, a) \in R.

Since a was arbitrary, R is reflexive.

"$a < b$" on $\mathbb{R}$ — a non-example

Reflexivity check. Let a \in \mathbb{R} be arbitrary. For (a, a) \in R we need a < a. But no real number is strictly less than itself. So (a, a) \notin R.

In fact we can say more: for every a, (a, a) \notin R, which means R is irreflexive (the opposite of reflexive).

Why: to disprove reflexivity we do not need to check every element either — exhibiting a single a with (a, a) \notin R already breaks the "for every" claim. Here every a fails, which is the strongest possible kind of non-reflexivity.

What the interactive below shows

Set size:

Slide to change the claimed size of a set $A$ on which we want to prove reflexivity. For small finite sizes you can enumerate the self-loops $(a,a)$ one by one. As the size grows the list becomes absurd — by the time you reach $\infty$, the only feasible argument is "let $a \in A$ be arbitrary" and check once. The canvas shows the self-loops that would need to be verified by enumeration; the recommended technique flips once the list gets unwieldy.

The takeaway

You never need to check every element of an infinite set. You let a single symbol a stand for "any element," argue once, and the quantifier does the rest. Reflexivity proofs on \mathbb{R}, \mathbb{Z}, \mathbb{Q}, \mathbb{N} are typically three lines: take an arbitrary a, apply the rule defining R to the pair (a, a), conclude.

If you ever find yourself writing "check a = 1… check a = 2… check a = 3…" on an infinite set, stop. Replace the list with "let a be arbitrary" and do the check once, symbolically.