Determine whether the following statement is true or false. Justify your answer: For all sets $A$ and $B$, $(A - B) \cup (A \cap B) = A$.

Determine whether the following statement is true – Class 11 Mathematics

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Concept understanding — Set Difference

Set Difference

The idea in plain words

Imagine two groups of students: those who play cricket ( $A$ ) and those who play football ( $B$ ). The set difference $A - B$ (also written $A \setminus B$ ) answers one specific question: "Who plays cricket but NOT football?" You start with everything in $A$ , then remove whatever also happens to be in $B$ .

Set difference is a one-way street: $A - B$ keeps only what's uniquely in $A$ . It has nothing to do with what's uniquely in $B$ .

The precise definition

For two sets $A$ and $B$ :

A - B = \{ x \mid x \in A \text{ and } x \notin B \}

Read as: "the set of all $x$ such that $x$ is in $A$ but $x$ is not in $B$ ."

Worked example

Let:

A = \{1, 2, 3, 4, 5\}, \qquad B = \{3, 4, 5, 6, 7\}

Step 1: Go through each element of $A$ .

Step 2: Keep it only if it is NOT also in $B$ .

$1 \in A$ , $1 \notin B$ → keep
$2 \in A$ , $2 \notin B$ → keep
$3 \in A$ , $3 \in B$ → remove
$4 \in A$ , $4 \in B$ → remove
$5 \in A$ , $5 \in B$ → remove

A - B = \{1, 2\}

Now compute the other direction:

B - A = \{6, 7\}

Notice $A - B \neq B - A$ — set difference is not commutative.

Key properties

Property	Statement
Not commutative	$A - B \neq B - A$ in general
Difference with itself	$A - A = \emptyset$
Difference with empty set	$A - \emptyset = A$ , and $\emptyset - A = \emptyset$
Difference with universal set	$U - A = A^c$ (the complement of $A$ )
Disjoint sets	If $A \cap B = \emptyset$ , then $A - B = A$

The last property is worth pausing on: if two sets share nothing in common, subtracting one from the other changes nothing — there was nothing to remove.

Set difference vs. complement — the classic mix-up

Students frequently confuse $A - B$ with $A^c$ (complement of $A$ ). The difference is what you're comparing against:

Complement $A^c$ is always relative to the universal set $U$ : everything outside $A$ .
Difference $A - B$ is relative to whatever second set you name: everything in $A$ that isn't in $B$ .

In fact, complement is just a special case: $A^c = U - A$ .

Set difference vs. symmetric difference …

Why this formula?

Let's break down the definition of a set — not as a formula to memorise, but as a fundamental idea that underpins all of mathematics.

1. What is a Set? (The Core Idea)

A set is a well-defined collection of distinct objects.

The "why" here is about clarity and precision — we need to know exactly what belongs and what does not.

Well-defined: For any object, we can say yes or no — no ambiguity.
Distinct: No duplicates — each object appears only once.

Why? Because if we couldn't decide membership, we couldn't do any logical operations. Sets are the building blocks of all mathematical structures.

2. The Key "Formula": Set-Builder Notation

The most common way to define a set is:

S = \{ x \mid P(x) \}

This reads: "S is the set of all objects $x$ such that property $P(x)$ is true."

Why does this work?

$x$ is a placeholder for any object.
$P(x)$ is a logical condition (a predicate) that is either true or false for each $x$ .
The vertical bar $\mid$ means "such that".

Example:

A = \{ n \mid n \in \mathbb{N}, n \text{ is even} \}

Here, $P(n)$ is " $n$ is a natural number and $n$ is even".

Only those $n$ that satisfy both conditions are included.

Why this form? It avoids listing infinitely many elements. It gives a rule — a decision procedure — for membership.

3. The Two Fundamental Properties (Axioms)

Every set definition relies on two intuitive truths:

(a) Extensionality — Two sets are equal if they have the same elements.

A = B \iff (\forall x)(x \in A \iff x \in B)

Why? A set is completely determined by its members. There is no other hidden property.

If you know what's inside, you know the set.

(b) Membership — The only relation is $\in$ (belongs to).

x \in S \quad \text{or} \quad x \notin S

Why? Because a set is just a container. The only question we can ask is: "Is this object inside?"

4. Why Can't We Just List Everything?

For small sets, listing works:

\{1, 2, 3\}

But for infinite sets (like all natural numbers), listing is impossible.

Set-builder notation solves this by giving a rule instead of a list.

Example:

\mathbb{N} = \{ n \mid n \text{ is a positive integer} \}

This is not a formula to memorise — it's a definition by property.

5. The "Empty Set" — Why It Exists

The empty set $\emptyset$ (or $\{\}$ ) is the set with no elements. …

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The statement is true because the set difference $A - B$ (elements in $A$ but not $B$ ) and the intersection $A \cap B$ (elements in both $A$ and $B$ ) together form a partition of set $A$ .

When we talk about sets, understanding how different operations combine or separate elements is crucial. The statement $(A - B) \cup (A \cap B) = A$ essentially asks if we can reconstruct set $A$ by combining two specific parts:

The elements that are in $A$ but not in $B$ (this is $A - B$ ).
The elements that are in $A$ and also in $B$ (this is $A \cap B$ ).

Imagine set $A$ as a group of students. Set $B$ is another group of students.

$A - B$ would be the students who are in group $A$ but not in group $B$ .
$A \cap B$ would be the students who are in group $A$ and also in group $B$ .

If you take all the students who are in group $A$ but not in group $B$ , and then you add all the students who are in group $A$ and also in group $B$ , what do you get? You get all the students who are in group $A$ . This intuitive understanding suggests the statement is true.

To formally prove this, we need to show that every element in $(A - B) \cup (A \cap B)$ is also in $A$ , and conversely, every element in $A$ is also in $(A - B) \cup (A \cap B)$ . This is the standard method for proving set equality.

Understanding the Definitions: Let's first define the terms involved:
- The set difference $A - B$ (read as "A minus B") consists of all elements that are in $A$ but not in $B$ .

A - B = \{x \mid x \in A \text{ and } x \notin B\}

*   The intersection $A \cap B$ (read as "A intersection B") consists of all elements that are common to both $A$ and $B$.

A \cap B = \{x \mid x \in A \text{ and } x \in B\}

*   The union $X \cup Y$ (read as "X union Y") consists of all elements that are in $X$ or in $Y$ (or both).

X \cup Y = \{x \mid x \in X \text{ or } x \in Y\}

Proving $(A - B) \cup (A \cap B) \subseteq A$ :

We need to show that if an element $x$ belongs to $(A - B) \cup (A \cap B)$ , then it must also belong to $A$ .

Let $x \in (A - B) \cup (A \cap B)$ .

By the definition of union, this means $x \in (A - B)$ or $x \in (A \cap B)$ .
- Case 1: $x \in (A - B)$
  
  By the definition of set difference, $x \in A$ and $x \notin B$ .
  
  From this, we can directly conclude that $x \in A$ .
- Case 2: $x \in (A \cap B)$
  
  By the definition of intersection, $x \in A$ and $x \in B$ .
  
  From this, we can directly conclude that $x \in A$ .
In both possible cases, we find that $x \in A$ . Therefore, any element in $(A - B) \cup (A \cap B)$ must also be in $A$ . This proves that $(A - B) \cup (A \cap B) \subseteq A$ .
Proving $A \subseteq (A - B) \cup (A \cap B)$ :

We need to show that if an element $x$ belongs to $A$ , then it must also belong to $(A - B) \cup (A \cap B)$ .

Let $x \in A$ .

For any element $x$ and any set $B$ , there are only two possibilities regarding $x$ 's membership in $B$ : either $x \in B$ or $x \notin B$ . These two possibilities are mutually exclusive and exhaustive. …

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Q.Determine whether the following statement is true or false. Justify your answer: For all sets AAA and BBB, (A−B)∪(A∩B)=A(A - B) \cup (A \cap B) = A(A−B)∪(A∩B)=A.

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Q.Determine whether the following statement is true or false. Justify your answer: For all sets $A$ and $B$ , $(A - B) \cup (A \cap B) = A$ .

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