When one bubble sits wholly inside another — subset, proper subset, equality, and the 2ⁿ subsets hiding in a set.
Sets don't live in isolation; the moment we have two of them, we compare them. The simplest comparison of all is containment: if every element of one set is also in another, the first is a subset of the second — picture the small circle nested inside the big one. Tighten that idea and you get a proper subset; turn it both ways at once and you get equality — two sets are the same exactly when each is a subset of the other. There is one quiet surprise hiding in the corner: the empty set ∅ is a subset of every set, for a reason that sounds like a riddle. And finally we count, and a power of two falls out: an n-element set secretly contains 2ⁿ subsets, because each element faces one yes/no choice — take it or leave it.
Start with the cleanest relationship two sets can have. We say A is a subset of B, written A ⊆ B, when every element of A is also an element of B. Read the symbol aloud as "A is a subset of B" or "A is contained in B." Picture the circle for A drawn entirely inside the circle for B: nothing of A pokes out.
Some examples you can check by eye. {2, 4} ⊆ {1, 2, 3, 4, 5} — both 2 and 4 are on the right. The even numbers are a subset of the integers, {…, −2, 0, 2, 4, …} ⊆ ℤ. And the famous tower of number families from Lesson 19.1 is one long chain of subsets: ℕ ⊆ ℤ ⊆ ℚ ⊆ ℝ.
A ⊆ B means: for every x, if x ∈ A then x ∈ B. To prove it, walk through A's elements and confirm each one is in B. To disprove it, you need just one element of A that is missing from B — a single counterexample is enough.
That asymmetry is worth pausing on. Proving a subset claim is patient work — you must check all of A. But refuting one is fast: find one renegade element of A that isn't in B and you're done. For instance, {1, 2, 5} ⊆ {1, 2, 3, 4} is false, because 5 ∈ {1,2,5} but 5 ∉ {1,2,3,4}. One witness settles it.
The same relationship read the other way around has its own name: a superset. B ⊇ A ("B contains A") says exactly what A ⊆ B says — the big circle swallowing the small one, seen from the big circle's side.
Keep two relationships apart. ∈ links an element to a set: 2 ∈ {1, 2, 3} ✓. ⊆ links a set to a set: {2} ⊆ {1, 2, 3} ✓ — but 2 ⊆ {1,2,3} and {2} ∈ {1,2,3} are both nonsense. The braces make the difference: 2 is a number, {2} is a one-element set.
Here is a fact that surprises people the first time: every set is a subset of itself, A ⊆ A. Test it against the definition — is every element of A also in A? Of course; that's just saying each element equals itself. So A ⊆ A is always true, the same way every number satisfies x ≤ x.
But often we want to say more than "inside or equal" — we want "strictly inside, genuinely smaller." When A ⊆ B and B has at least one extra element that A lacks (so A ≠ B), we call A a proper subset and write A ⊊ B. Thus {2, 4} ⊊ {1, 2, 3, 4, 5} — a proper subset, because 1, 3, and 5 are left over. But {1, 2, 3} ⊆ {1, 2, 3} is not proper: nothing is left over, the two sets coincide.
The pair ⊆ and ⊊ works just like ≤ and < for numbers. A ⊆ B permits A = B; A ⊊ B rules it out by demanding B be strictly larger. So A ⊆ A is always true, but A ⊊ A is always false.
When are two sets equal? The honest answer from Lesson 19.1 is: when they have exactly the same members — order and repeats never count, so {1, 2, 2} = {2, 1}. But there is a second, more powerful way to say the same thing, and it is the standard tool for proving equality:
A = B exactly when A ⊆ B and B ⊆ A. Each set is a subset of the other, so neither has anything the other lacks — they must be the very same set. This "containment both ways" is how mathematicians prove that two differently-described sets are identical.
An example shows why it earns its keep. Claim: { x ∈ ℤ | x² = 4 } = {−2, 2}. To settle it, prove both directions. (⊆) Any integer x with x² = 4 must be −2 or 2, so the left set is contained in {−2, 2}. (⊇) Both −2 and 2 are integers whose square is 4, so {−2, 2} is contained in the left set. Both containments hold ⇒ the sets are equal. ✓
Because order and repeats are invisible to a set, two lists that look different can name one set: {3, 1, 2} = {1, 2, 3} and {2, 2, 3} = {2, 3}. Equality is about which members, never about how you wrote them down.
Now the riddle. The empty set ∅ — the set with no elements at all, met in Lesson 19.2 — is a subset of every set: ∅ ⊆ A, no matter what A is. Even ∅ ⊆ ∅. How can the bubble that holds nothing fit inside everything?
Run the definition and see. To say ∅ ⊆ A is false, we would have to exhibit an element of ∅ that is missing from A. But ∅ has no elements — so there is no such witness to produce, and the claim cannot fail. A statement that can't fail is true; mathematicians call this vacuously true. The condition "every element of ∅ is in A" is satisfied for free, because there are no elements to check.
Every set A comes with two guaranteed subsets, no matter what's inside it: ∅ ⊆ A (the empty set) and A ⊆ A (the set itself). They sit at the two extremes — the smallest possible subset and the largest. For a non-empty A these are two different subsets: ∅ ⊊ A is a proper subset (strictly smaller), while A ⊆ A is never proper. When A = ∅ the two coincide — ∅'s only subset is ∅ itself, and ∅ ⊊ ∅ is false.
How many subsets does a set with n elements have? Don't try to list them blindly; reason element by element. Build a subset by walking down A's members and, for each one, making a single decision: is this element in my subset, or out? Two choices per element, and the choices are completely independent of one another.
So the total number of subsets is 2 × 2 × ⋯ × 2 — one factor of 2 for each of the n elements — which is 2ⁿ. For A = {a, b, c}, that's 2³ = 8 subsets. Here they all are, organized by size:
| size 0 | size 1 | size 2 | size 3 |
|---|---|---|---|
| ∅ | {a} {b} {c} |
{a, b} {a, c} {b, c} |
{a, b, c} |
| 1 | 3 | 3 | 1 |
That's 1 + 3 + 3 + 1 = 8 ✓. The collection of all subsets of A is itself a set, called the power set of A. Notice the bookkeeping: of the 2ⁿ subsets, exactly one is ∅ and exactly one is A itself. Drop A and you are left with 2ⁿ − 1 proper subsets; drop both ∅ and A and you have 2ⁿ − 2 that are proper and nonempty.
The 2ⁿ count is the multiplication principle in disguise: n independent choices, two options each, multiply to give the total. The doubling is relentless — each new element doubles the count, because it can be tacked onto every existing subset or left off it.
Subset. A ⊆ B ⟺ every element of A is in B. One missing element disproves it; B ⊇ A says the same thing backwards.
Proper subset. A ⊊ B ⟺ A ⊆ B and A ≠ B. Like ≤ vs <: ⊆ allows equality, ⊊ forbids it. Always A ⊆ A.
Equality. A = B ⟺ A ⊆ B and B ⊆ A — containment both ways is the standard proof. Order and repeats never matter.
Empty set. ∅ ⊆ A for every A (vacuously true — no element can be missing). So every set has ∅ and itself as subsets.
Counting. An n-element set has 2ⁿ subsets (each element: in or out). Of these, 2ⁿ − 1 are proper, 2ⁿ − 2 are proper and nonempty.
Is {1, 3} ⊆ {1, 2, 3, 4}? Explain in one line.
Yes. Both 1 and 3 are in {1, 2, 3, 4}, so every element of {1, 3} is in {1, 2, 3, 4} ⇒ {1, 3} ⊆ {1, 2, 3, 4}. (It is in fact a proper subset, since 2 and 4 are left over.)
Give a counterexample showing {1, 2, 5} ⊆ {1, 2, 3, 4} is false.
5 ∈ {1, 2, 5} but 5 ∉ {1, 2, 3, 4}. One element of the first set is missing from the second, so containment fails — the subset claim is false.
True or false: {2, 2, 3} = {2, 3}?
True. Repeats don't count in a set, so {2, 2, 3} names the same two members as {2, 3}: they have identical elements, hence are equal.
How many subsets does {a, b, c, d} have? How many are proper?
Each of the 4 elements is in or out, independently: 2⁴ = 16 subsets. Dropping the whole set itself leaves 16 − 1 = 15 proper subsets.
True or false: ∅ ⊆ {7}? Why?
True. The empty set is a subset of every set: to fail, ∅ would need an element not in {7}, but ∅ has no elements, so it cannot fail (vacuously true).
List all subsets of {1, 2}.
∅, {1}, {2}, {1, 2} — four of them, matching 2² = 4. (Don't forget ∅ and the whole set.)
Six questions to lock it in. Tap the answer you think is right.
The big idea. Comparison is the heart of this lesson: a subset is the "is-contained-in" relation, and equality is just that relation run both ways. The picture — a small circle inside a big one — is the whole concept, and students should narrate it: "every element of A is in B, so A ⊆ B." The 2ⁿ count is the first taste of the multiplication principle, made vivid by the in/out switch on each element.
The misconception to watch. The single most common error is confusing ∈ (element) with ⊆ (subset). Drill the distinction out loud: 2 is an element of {1,2,3} (2 ∈), but {2} is a subset of {1,2,3} ({2} ⊆). The braces are not decoration — 2 and {2} are different objects. Two more traps: forgetting ∅ and A itself when listing or counting subsets (so {1,2} has four, not two), and doubting that A ⊆ A or ∅ ⊆ A can be true. The empty-set fact rewards a slow reading of "vacuously true."
Common Core. This lesson supports the high-school convention for set relations (subset, proper subset, equality) and the counting principle behind 2ⁿ subsets (S-CP / the fundamental counting principle). The symbols are most of the difficulty — insist students read every one aloud: "is a subset of," "is an element of," "the empty set."