Element or subset?

Nested boxes for A = {1, 2, {3, 4}}: an element is a chip inside the outer box, a subset is a dashed loop drawn round whole chips; the sim judges each statement as ∈ or ⊂ and shows why.

Build the one true subset statement that uses {3, 4}: lasso the inner box so the candidate is {{3, 4}} with ⊂ selected (and confirm {3, 4} ⊂ A is rejected).

Writing {3, 4} ⊂ A because {3, 4} is 'in' A; treating {φ} as empty.

• A ⊂ B (A is a subset of B) means each member of A also belongs to B; symbolically, a ∈ A ⇒ a ∈ B. If even one member of A lies outside B, then A ⊄ B.
• Every set is a subset of itself, A ⊂ A, and the empty set is a subset of every set, since it has no element that could fail the test.
• A ⊂ B and B ⊂ A together hold exactly when A = B; proving both inclusions is the standard way to prove two sets equal.
• If A ⊂ B and A ≠ B, A is a proper subset of B and B is a superset of A. A set with exactly one element, such as {a}, is a singleton.
• ∈ joins an element to a set; ⊂ joins a set to a set. With A = {1, 2, {3, 4}}: {3, 4} ∈ A and {{3, 4}} ⊂ A are true, but {3, 4} ⊂ A is false because 3 ∉ A.
• A ∈ B and B ⊂ C do not force A ⊂ C: with A = {5}, B = {{5}, 6} and C = {{5}, 6, 7}, A is an element of B and B ⊂ C, yet A ⊄ C because 5 ∉ C.
• Listing subsets by size shows that each new element doubles the count: {−1, 0, 1} has 1 + 3 + 3 + 1 = 8 subsets, and in general a set with n elements has 2ⁿ subsets, counting φ and the set itself (a counting result derived from the listing, not a separate NCERT statement).

Teaching this idea? Put the simulation on your own page or LMS with this code.