A student claims: 'The empty set ∅ cannot be a subset of {1, 2, 3} because ∅ contains no elements, so it has nothing in common with {1, 2, 3}.' What is wrong with this reasoning?
AThe student is correct — ∅ is only a subset of itself
BThe subset relation requires at least one shared element, but ∅ satisfies a stricter version that counts as a subset anyway
CThe definition of A ⊆ B requires every element of A to be in B; since ∅ has no elements, this condition is vacuously true — ∅ is a subset of every set
D∅ ⊆ {1, 2, 3} is true, but only because ∅ is a special case defined by convention, not by the general subset definition
The subset relation A ⊆ B means: for every x, if x ∈ A then x ∈ B. When A = ∅, there are no elements x to check — the condition is vacuously satisfied. This is not a convention or special case; it follows directly from the definition. The student's error is confusing 'has elements in common with' (intersection) with the subset relation, which is purely about containment direction.
Question 2 Multiple Choice
A student is asked whether {1, 2, 3} is a proper subset of {1, 2, 3}. They answer 'yes, because it is clearly contained within it.' What is wrong?
ANothing — a set is always a proper subset of itself
BA proper subset requires strict containment: A ⊂ B means A ⊆ B and A ≠ B. Since the sets are equal, {1,2,3} is a subset but NOT a proper subset of itself
CThe question is ill-formed because a set cannot be compared to itself
DA set is neither a subset nor a proper subset of itself
A proper subset adds the condition A ≠ B. Every set is a subset of itself (A ⊆ A), but no set is a proper subset of itself (A ⊂ A is always false). The student confused ⊆ (subset, permits equality) with ⊂ (proper subset, requires strict containment). The distinction mirrors < vs. ≤ for numbers: both express ordering, but only one permits equality.
Question 3 True / False
If A ⊆ B and B ⊆ A, then A = B.
TTrue
FFalse
Answer: True
This is the antisymmetry property of the subset relation, and it is the standard proof technique for set equality: to show two sets are equal, show each is a subset of the other. If A ⊆ B, every element of A is in B. If B ⊆ A, every element of B is in A. Together, A and B contain exactly the same elements, so by the axiom of extensionality, A = B.
Question 4 True / False
Most set is a proper subset of itself.
TTrue
FFalse
Answer: False
A proper subset A ⊂ B requires A ⊆ B and A ≠ B. Since A = A (every set equals itself), the condition A ≠ B fails when B = A. Therefore A ⊂ A is always false — no set is a proper subset of itself. Every set IS a subset of itself (A ⊆ A, which follows from reflexivity), but the 'proper' qualifier excludes the case of equality.
Question 5 Short Answer
Why is the empty set a subset of every set, including itself? Explain using the definition of subset.
Think about your answer, then reveal below.
Model answer: A ⊆ B is defined as: for every element x, if x ∈ A then x ∈ B. When A = ∅, there are no elements x such that x ∈ ∅, so the conditional 'if x ∈ ∅ then x ∈ B' is never triggered — it is vacuously true for any set B, including ∅ itself. The empty set satisfies the subset definition trivially because it has no elements that could fail to be in B.
Vacuous truth is a precise logical concept: a universal statement 'for all x, P(x) → Q(x)' is true when P(x) is false for every x, because no counterexample exists. The empty set has no elements, so no element can witness a failure of the subset condition. This is not a coincidence or a convention — it is what the definition logically entails.