# Set Theory: Membership and Operations

Philosophy I · Logic · https://tryals.app/learn/philosophy-i/set-theory-membership-and-operations

## Collections, Taken Seriously

A **set** is a collection of distinct objects, its **members**. Everything follows from two decisions: a set is fixed entirely by *which* objects belong to it, the **axiom of extensionality**, and it has no order and no repetition. So $\{1, 2\}$, $\{2, 1\}$ and $\{1, 2, 2\}$ are all the same set.

Sets are specified by listing, $\{1, 2, 3\}$, or by a condition, $\{x : x \text{ is even}\}$. The **empty set** $\emptyset$ has no members and is unique, since any two memberless sets have the same members and are therefore identical.

Two relations are constantly confused. **Membership** $a \in A$ holds between an object and a set. **Inclusion** $A \subseteq B$ holds between two sets, when every member of $A$ is also a member of $B$. They are different relations, and $\{1\} \in \{\{1\}\}$ while $\{1\} \subseteq \{1, 2\}$.

The operations combine sets:

| Operation | Notation | Members |
|---|---|---|
| Union | $A \cup B$ | in $A$, or in $B$, or both |
| Intersection | $A \cap B$ | in both |
| Difference | $A \setminus B$ | in $A$ but not $B$ |
| Complement | $A^{c}$ | in the universe but not $A$ |

The parallel with logic is exact and not a coincidence: union behaves like disjunction, intersection like conjunction, complement like negation. De Morgan's laws hold in both, in the same form.

Counting gives the **inclusion-exclusion** principle:

$$|A \cup B| = |A| + |B| - |A \cap B|$$

The subtraction is the whole content: adding the two sizes counts the overlap twice, so it must be removed once. For 30 students taking logic, 25 taking ethics and 10 taking both, the total is $30 + 25 - 10 = 45$, not 55.

> **Common pitfall:** confusing $\in$ with $\subseteq$. The empty set is a **subset** of every set — vacuously, since it has no member that could fail to belong — but it is a **member** only of sets that explicitly contain it. $\emptyset \subseteq \{1\}$ is true; $\emptyset \in \{1\}$ is false.

## Practice questions

7 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. A class has 18 students taking logic, 14 taking metaphysics, and 6 taking both. How many students take at least one of the two?

**Answer:** 26

**Why:** $18 + 14 - 6 = 26$. Answering 32 is the standard error, and it double-counts the six students who appear in both lists.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/a-class-has-18-students-taking-logic-14-taking-metaphysics-and-6

### 2. Which statements about the empty set are correct?

A. It is a member of every set
B. It is unique
C. It is a subset of every set
D. It has no members

**Answer:** B. It is unique; C. It is a subset of every set; D. It has no members

**Why:** The empty set is a subset of every set vacuously, it has no member that could fail to belong. It is a *member* only of sets that explicitly contain it: $\emptyset \subseteq \{1\}$ is true while $\emptyset \in \{1\}$ is false.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/which-statements-about-the-empty-set-are-correct

### 3. De Morgan’s laws hold in the same form for sets as for propositions.

**Answer:** True

**Why:** True, the complement of an intersection is the union of the complements, exactly as the negation of a conjunction is the disjunction of the negations. The parallel is structural, not notational: both are descriptions of one underlying algebra.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/de-morgans-laws-hold-in-the-same-form-for-sets-as-for-propositions

### 4. Sort each claim by whether it uses membership or inclusion correctly.

**Answer:**

- Correct use of membership: The number 2 belongs to the set of even numbers, The set containing 1 belongs to the set whose only member is that set
- Correct use of inclusion: The set containing 1 is contained in the set containing 1 and 2, The empty set is contained in every set

**Why:** Membership is object-to-set and inclusion is set-to-set. A set can be a member of one collection and a subset of another, which is exactly why the two symbols must be kept apart.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/sort-each-claim-by-whether-it-uses-membership-or-inclusion-correctly

### 5. Arrange these sets in order of increasing size.

**Answer:**

1. The empty set
2. The set of even prime numbers
3. The set of vowels in the English alphabet
4. The set of days in a week

**Why:** Zero, one, five, seven. The even primes are a set with exactly one member, 2, which is a useful reminder that a condition can pick out a single object without that being a coincidence.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/arrange-these-sets-in-order-of-increasing-size

### 6. The axiom of extensionality defines a set entirely by its members, without order or repetition. What consequence does this principle have for how we determine whether two separately described sets are identical?

A. They are identical if they contain the same total count of items
B. They are identical only if both are specified by identical rules
C. They are identical if and only if each is a subset of the other
D. They are identical only if their elements appear in matching order

**Answer:** C. They are identical if and only if each is a subset of the other

**Why:** Equivalence in extensionality ignores description methods or counts; it requires mutual inclusion ($A \subseteq B$ and $B \subseteq A$). Confusing intensional definitions or cardinality with set identity overlooks that distinct predicates can pick out the exact same collection of objects.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/the-axiom-of-extensionality-defines-a-set-entirely-by-its-members

### 7. Match each notation to what it expresses.

**Answer:**

- a is a member of A → An object belongs to a set
- A is a subset of B → Every member of A is a member of B
- A union B → Everything in either set
- A intersect B → Everything in both sets

**Why:** Relations either hold or do not; operations build a new set from old ones. Keeping the distinction clear matters, since a relation cannot be a member of anything but the result of an operation can.

Page: https://tryals.app/practice/philosophy-i/set-theory-membership-and-operations/match-each-notation-to-what-it-expresses