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Closure (Sets) — Definition, Formula & Examples

Closure is the property that a set has when performing a specific operation on any elements of the set always produces a result that is also in the set. For example, the integers are closed under addition because adding any two integers always gives another integer.

A set SS is said to be closed under a binary operation * if for every a,bSa, b \in S, the result abSa * b \in S. More generally, SS is closed under an nn-ary operation ff if f(a1,a2,,an)Sf(a_1, a_2, \ldots, a_n) \in S whenever all aiSa_i \in S.

How It Works

To test whether a set is closed under an operation, you check whether every possible application of that operation to elements in the set yields a result that stays in the set. A single counterexample — one pair of elements whose result falls outside the set — is enough to prove the set is not closed. You do not need to check every pair if you can find a general argument covering all cases.

Worked Example

Problem: Determine whether the set S={0,1,2,3,4,}S = \{0, 1, 2, 3, 4, \ldots\} (the non-negative integers) is closed under subtraction.
Pick elements from S: Choose two elements from SS, say a=2a = 2 and b=5b = 5.
a=2,  b=5a = 2,\; b = 5
Apply the operation: Compute aba - b.
25=32 - 5 = -3
Check membership: Is 3-3 in SS? No — the set SS contains only non-negative integers, and 3-3 is negative.
3S-3 \notin S
Answer: The set of non-negative integers is not closed under subtraction, because the counterexample 25=32 - 5 = -3 produces a result outside the set.

Why It Matters

Closure is a foundational requirement in abstract algebra: a group, ring, or field must be closed under its operations. In linear algebra, verifying closure under addition and scalar multiplication is the first step when proving something is a subspace.

Common Mistakes

Mistake: Testing only a few examples and concluding the set is closed.
Correction: A few successful examples do not prove closure. You need a general argument that covers all elements, or a single counterexample to disprove it.