Operations on Sets and Statements: Union/Intersection and “Or”/“And”

Why this matters

In the last two posts, we learned two big ideas:

• Sets let us talk about membership (who belongs, who doesn’t).

• Variables and quantifiers let us turn templates into true-or-false statements.

This post connects those ideas. When you combine sets, you are often doing the same logical work as combining statements with and and or.

Two core set operations

Suppose you have two sets, R and S.

Intersection: “in both”

The intersection of R and S is written:

R ∩ S

It means: all elements that belong to both sets.

In plain English:

x ∈ R ∩ S means "x is in R and x is in S."

Union: “in at least one”

The union of R and S is written:

R ∪ S

It means: all elements that belong to at least one of the sets.

In plain English:

x ∈ R ∪ S means "x is in R or x is in S (or both)."

Venn diagrams as a visual check

A Venn diagram can help you see these operations quickly:

• R ∩ S: shade only the overlap

• R ∪ S: shade everything in either circle

How this connects to open sentences (solution sets)

Set operations become especially useful when sets come from conditions.

Here’s the typical situation:

• You have a domain (like integers, or real numbers).

• You have two open sentences (conditions).

• Each condition produces a solution set.

• Then you want “solutions that satisfy both” or “solutions that satisfy at least one.”

Example

Let the domain be the integers.

Define two sets using conditions:

• A = {x ∈ ℤ : x² = 9}

• B = {x ∈ ℤ : x − 1 = 0}

Find each set:

• x² = 9 has integer solutions x = -3 and x = 3, so A = {-3, 3}

• x - 1 = 0 has integer solution x = 1, so B = {1}

Now combine them:

• Intersection: A ∩ B = {-3, 3} ∩ {1} = ∅ (No integer makes both conditions true.)

• Union: A ∪ B = {-3, 3} ∪ {1} = {-3, 1, 3} (Integers that satisfy at least one condition.)

Notice the difference in meaning:

• A ∩ B corresponds to “Condition 1 and Condition 2”

• A ∪ B corresponds to “Condition 1 or Condition 2”

Operations on statements: “and” and “or”

To talk about combining statements cleanly, mathematicians often use letters:

• Let p represent one statement.

• Let q represent another statement.

Conjunction: p and q

The statement “p and q” is called a conjunction.

It is true only when both p and q are true.

Example:

• p: “5 is odd.” (true)

• q: “5 is divisible by 4.” (false)

Then:

• p and q is false (because q is false)

Disjunction: p or q

The statement “p or q” is called a disjunction.

In mathematics, “or” is usually inclusive:

• it is true if p is true

• or q is true

• or both are true

The disjunction is false only when both are false.

Example:

• p: “10 is even.” (true)

• q: “10 is prime.” (false)

Then:

• p or q is true (because at least one part is true)

Important note: everyday “or” vs mathematical “or”

In everyday conversation, “or” sometimes suggests “exactly one”:

• “You can have soup or salad.”

Often this implies you can’t have both.

In mathematics, unless the context clearly forces a different meaning, “or” means:

one or the other or both

So if both parts happen to be true, the disjunction is still true.

Example:

• "3 < 7 or 3 is positive." Both parts are true, and the whole statement is true.

The bridge between set operations and statement operations

This is the key connection:

Intersection matches “and”

x ∈ R ∩ S ⟺ (x ∈ R) and (x ∈ S)

Union matches “or”

x ∈ R ∪ S ⟺ (x ∈ R) or (x ∈ S)

So when you shade a Venn diagram for R ∩ S or R ∪ S, you are basically drawing the truth behavior of “and” and “or,” but for membership instead of sentences.

Summary

• R ∩ S means “in both” and matches logical and.

• R ∪ S means “in at least one” and matches logical or (inclusive).

• When sets are defined by conditions, union/intersection let you combine conditions cleanly.

• Venn diagrams are a useful way to check your thinking quickly.

Mini practice (with answers)

1) Set operations

Let R = {1, 2, 3, 4} and S = {3, 4, 5, 6}.

a) Find R ∩ S. b) Find R ∪ S.

Answers: a) {3, 4} b) {1, 2, 3, 4, 5, 6}

2) Condition sets over integers

Let the domain be integers and define:

• A = {x ∈ ℤ : x² = 16}

• B = {x ∈ ℤ : x is even}

a) Find A. b) Describe A ∩ B. c) Describe A ∪ B.

Answers: a) A = {-4, 4} b) A ∩ B = {-4, 4} (both are even, so all of A stays) c) A ∪ B = all even integers (since A is already inside the even integers)

3) Logic “or” check

Let:

• p: “9 is a multiple of 3.” (true)

• q: “9 is prime.” (false)

Is p or q true or false?

Answer: True, because at least one part is true.

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