Grade 6

Sets and set operations

A set is a well-defined collection of objects, its elements; $x \in A$ says $x$ belongs to $A$ . One set is a subset of another, $A \subseteq B$ , when every element of $A$ is also in $B$ . From two sets we build new ones: the union $A \cup B$ (elements in $A$ or in $B$ ), the intersection $A \cap B$ (elements in both), and the difference $A \setminus B$ (in $A$ but not $B$ ). These operations obey algebra-like laws such as the distributive law $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ , and counting elements of a union follows the inclusion-exclusion rule $|A \cup B| = |A| + |B| - |A \cap B|$ , both proved rigorously here and applied to survey data and search-engine filters.

IntuitionBoxes of objects: what belongs where

Imagine two overlapping boxes of toys: box $A$ holds building blocks, box $B$ holds cars. Some toys are block-cars and sit in both boxes at once. Asking " $x \in A$ ?" is asking whether a specific toy $x$ is inside box $A$ . Asking " $A \subseteq B$ ?" is asking whether every single toy in box $A$ is also somewhere in box $B$ . Combining the two boxes into one pile gives the union $A \cup B$ ; keeping only toys that are in both boxes gives the intersection $A \cap B$ ; taking box $A$ and throwing out anything also in box $B$ gives the difference $A \setminus B$ . The network widget below lets you drag elements between two circles and watch the four operations recompute live.

Venn diagram of three sets

A, B, C

showing union, pairwise intersections, and the central triple intersection

A\cap B\cap C

SchoolFormal definitions and a comparison table

Definition: Set, element, subset

A set $A$ is a collection of distinct objects; each object $x$ with $x \in A$ is called an element of $A$ . Set $A$ is a subset of set $B$ , written $A \subseteq B$ , when every element of $A$ is also an element of $B$ ; two sets are equal exactly when $A \subseteq B$ and $B \subseteq A$ hold simultaneously.

A \cup B = \{x : x \in A \lor x \in B\}

The union $A \cup B = \{x : x \in A \lor x \in B\}$ collects an element as soon as it belongs to at least one of the two sets. Dually, the intersection keeps only elements shared by both sets.

A \cap B = \{x : x \in A \land x \in B\}

Set operations on $A=\{1,2,3\}$ and $B=\{2,3,4\}$
Operation	Definition	Result
Union $A \cup B$	$A \cup B = \{x : x \in A \lor x \in B\}$	$\{1,2,3,4\}$
Intersection $A \cap B$	$A \cap B = \{x : x \in A \land x \in B\}$	$\{2,3\}$
Difference $A \setminus B$	$A \setminus B = \{x : x \in A \land x \notin B\}$	$\{1\}$
Subset $A \subseteq B$	every element of $A$ is in $B$	false ( $1 \notin B$ )

UndergraduateTwo key theorems, with full proofs

Theorem: Distributive law of intersection over union

For any three sets $A$ , $B$ , $C$ : $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

Why is it true?

Just as multiplication distributes over addition in ordinary algebra, intersection distributes over union in set algebra. This lets you rewrite a filter such as "in category $A$ , and (on sale $B$ or new arrival $C$ )" as two separately manageable filters combined by a union, which is exactly how database and search-engine query planners simplify boolean filters.

Proof

We prove set equality by showing both inclusions, using an arbitrary element $x_0$ (element-chasing).

( $\subseteq$ ) Suppose $x_0 \in A \cap (B \cup C)$ . By the definition of intersection this means $x_0 \in A$ and $x_0 \in B \cup C$ . By the definition of union, $x_0 \in B \cup C$ means $x_0 \in B$ or $x_0 \in C$ . Consider the two cases separately. Case 1: $x_0 \in B$ . Combined with $x_0 \in A$ , this gives $x_0 \in A \cap B$ . Case 2: $x_0 \in C$ . Combined with $x_0 \in A$ , this gives $x_0 \in A \cap C$ . In either case $x_0$ lies in $x_0 \in A \cap B$ or in $x_0 \in A \cap C$ , so $x_0 \in (A \cap B) \cup (A \cap C)$ .

( $\supseteq$ ) Suppose $x_0 \in (A \cap B) \cup (A \cap C)$ . By the definition of union this means $x_0 \in A \cap B$ or $x_0 \in A \cap C$ . If $x_0 \in A \cap B$ , then $x_0 \in A$ and $x_0 \in B$ , hence $x_0 \in B \cup C$ (since $x_0$ is in $B$ , it is in $B$ or $C$ ), so $x_0 \in A \cap (B \cup C)$ . If instead $x_0 \in A \cap C$ , then $x_0 \in A$ and $x_0 \in C$ , hence $x_0 \in B \cup C$ , so again $x_0 \in A \cap (B \cup C)$ .

Since every element of the left-hand side lies in the right-hand side and vice versa, the two sets contain exactly the same elements, so $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ . This finishes the proof.

Theorem: Two-set inclusion–exclusion

For any two finite sets $A$ and $B$ : $|A \cup B| = |A| + |B| - |A \cap B|$ , where $|A|$ , $|B|$ , $|A \cap B|$ , $|A \cup B|$ denote the number of elements of each set.

Why is it true?

Simply adding $|A|$ and $|B|$ double-counts every element that is in both sets, so subtracting $|A \cap B|$ once corrects the overcount. This is exactly the arithmetic behind reading a two-circle Venn diagram from survey data: people who like both coffee and tea must not be counted twice when asking how many people like at least one.

Proof

The key idea is to split $A \cup B$ into three pairwise-disjoint pieces and count each piece once.

First, partition each set using the other: $A = (A \setminus B) \cup (A \cap B)$ and $B = (B \setminus A) \cup (A \cap B)$ . In each equation the two pieces on the right are disjoint, because $(A\setminus B) \cap (A \cap B) = \varnothing$ (an element outside $B$ cannot simultaneously be inside $B$ ) and likewise $(B\setminus A) \cap (A \cap B) = \varnothing$ . Since a finite set's size equals the sum of the sizes of any partition into disjoint pieces, this gives $|A| = |A \setminus B| + |A \cap B|$ and $|B| = |B \setminus A| + |A \cap B|$ .

Next, observe that $A \cup B$ itself splits into three pairwise-disjoint pieces: $A \cup B = (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$ . Indeed $A\setminus B$ , $B\setminus A$ , $A\cap B$ are pairwise disjoint (an element in $A\setminus B$ is not in $B$ , hence not in $A \cap B$ or in $B \setminus A$ ; symmetrically for the others), and their union recovers exactly the elements that are in $A$ or in $B$ . Counting this partition gives $|A \cup B| = |A \setminus B| + |B \setminus A| + |A \cap B|$ .

Finally substitute: from $|A| = |A \setminus B| + |A \cap B|$ we get $|A \setminus B|$ $= |A| - |A\cap B|$ , and from $|B| = |B \setminus A| + |A \cap B|$ we get $|B \setminus A|$ $= |B| - |A\cap B|$ . Plugging both into $|A \cup B| = |A \setminus B| + |B \setminus A| + |A \cap B|$ gives $|A \cup B| = (|A| - |A\cap B|) + (|B| - |A\cap B|) + |A\cap B| = |A| + |B| - |A\cap B|$ , which is exactly $|A \cup B| = |A| + |B| - |A \cap B|$ . This finishes the proof.

UndergraduateReal-World Applications and Worked Examples

Set operations are the backbone of two everyday tools. In survey analysis, a two-circle Venn diagram of respondents who like product $A$ and product $B$ is read off using $|A \cup B| = |A| + |B| - |A \cap B|$ : total interest is not the raw sum, because people in $A \cap B$ would be counted twice. In search-engine and e-commerce filters, combining "category $A$ " AND "on sale $B$ " is literally the intersection $A \cap B$ , combining with OR is the union $A \cup B$ , and excluding a tag with NOT is the difference $A \setminus B$ ; the distributive law $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ is what lets a query planner rewrite " $A$ and ( $B$ or $C$ )" into "( $A$ and $B$ ) or ( $A$ and $C$ )" without changing the results.

Example: Reading a survey with inclusion–exclusion

A school surveys $120$ students. $70$ say they play soccer, $45$ say they play basketball, and $25$ say they play both. How many students play at least one of the two sports, and how many play neither?

Solution

Let $A$ be the set of students who play soccer and $B$ the set who play basketball, so $|A| = 70$ , $|B| = 45$ , $|A \cap B| = 25$ .

By the inclusion–exclusion theorem proved above, $|A \cup B| = |A| + |B| - |A \cap B|$ . Substituting the numbers: $|A \cup B| = 70 + 45 - 25 = 90$ . So $90$ students play at least one of the two sports.

Since the total surveyed is $120$ , the number playing neither sport is the complement of $A \cup B$ inside the whole group of $120$ students: $120 - 90 = 30$ .

Notice the common mistake of simply adding $70+45=115$ : this would double-count the $25$ students who play both, which is exactly why the theorem subtracts $|A \cap B|$ once.

Example: Simplifying a search filter with the distributive law

An online store's catalog is a set $U$ of products. Let $A$ be "in the Shoes category", $B$ be "on sale", and $C$ be "new arrival". A shopper's filter is "in Shoes, and (on sale or new arrival)", i.e. $A \cap (B \cup C)$ . Show this is the same product list as running two separate simpler filters and merging the results, and explain why a search engine prefers the second form.

Solution

By the distributive law proved above, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ . Applying it here with $A=$ Shoes, $B=$ on sale, $C=$ new arrival: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

The left-hand side $A \cap (B \cup C)$ describes a single combined filter: first build the set $B \cup C$ of everything on sale or new, then intersect with Shoes. The right-hand side describes two independent simple filters, "Shoes on sale" ( $A\cap B$ ) and "new Shoes" ( $A \cap C$ ), whose results are then merged with a union.

Because the theorem guarantees both sides list exactly the same products, a query planner is free to choose whichever execution is faster. Running two simple single-condition filters (each of which can use a fast pre-built index for one category) and merging with a union is usually cheaper than evaluating a nested "or inside an and" expression directly, so real search engines rewrite queries into the right-hand form before executing them.

Concretely, if Shoes $=\{1,2,3,4,5\}$ , on sale $=\{2,4,6\}$ , new arrival $=\{1,4,7\}$ , then $B\cup C=\{1,2,4,6,7\}$ and $A\cap(B\cup C)=\{1,2,4\}$ ; separately $A\cap B=\{2,4\}$ and $A\cap C=\{1,4\}$ , whose union is again $\{1,2,4\}$ , confirming the two computations agree.

Let $A=\{2,4,6,8\}$ . Which statement is correct?

A class has $30$ students; $18$ like math, $15$ like physics, and $9$ like both. How many like at least one of the two subjects?

A shopping site lets you filter "brand $A$ " OR "under 50 dollars ( $B$ )". Which set operation computes the results shown?

With $A=\{1,2,3\}$ , $B=\{2,3\}$ , $C=\{3,4\}$ , does $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ hold, and what is the common value?

References

Paul R. Halmos (1960). Naive Set Theory
Kenneth H. Rosen (2019). Discrete Mathematics and Its Applications