Using properties of sets, prove that for all sets $A$ and $B$, $A \cup (B - A) = A \cup B$.

> [!TLDR] > The key idea is that $B - A$ (elements in $B$ but not in $A$) together with $A$ covers everything in $A$ or $B$, so $A \cup (B - A) = A \cup B$. The proof uses the definition of set difference and the distributive law. ## Why This Works The expression $B - A$ (also written $B \setminus A$) means "all elements that are in $B$ but not in $A$." When you take the union of $A$ with $B - A$, you're collecting everything that is either in $A$ or in $B$ but not in $A$. But if an element is in $B$ and also in $A$, it's already captured by $A$ itself. So the union of $A$ and $B - A$ ends up being exactly the same as the union of $A$ and $B$ — nothing is lost, nothing extra is added. Let's prove this formally using set properties. ## Step-by-Step Proof 1. **Start with the left-hand side.** We want to show $A \cup (B - A) = A \cup B$. Recall the definition: $B - A = \{ x \mid x \in B \text{ and } x \notin A \}$. 2. **Rewrite $B - A$ using set operations.** The set difference can be expressed as an intersection with the complement: $$B - A = B \cap A^c$$ where $A^c$ is the complement of $A$ (relative to the universal set). So the left side becomes: $$A \cup (B \cap A^c)$$ 3. **Apply the distributive law.** The distributive law for sets says: $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$. Using this with $X = A$, $Y = B$, $Z = A^c$: $$A \cup (B \cap A^c) = (A \cup B) \cap (A \cup A^c)$$ 4. **Simplify $A \cup A^c$.** The union of a set and its complement is the universal set $U$ (everything under consideration): $$A \cup A^c = U$$ So we have: $$(A \cup B) \cap U$$ 5. **Intersection with the universal set.** …

Using properties of sets, prove that for all sets – 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 key idea is that $B - A$ (elements in $B$ but not in $A$ ) together with $A$ covers everything in $A$ or $B$ , so $A \cup (B - A) = A \cup B$ . The proof uses the definition of set difference and the distributive law.

Why This Works

The expression $B - A$ (also written $B \setminus A$ ) means "all elements that are in $B$ but not in $A$ ." When you take the union of $A$ with $B - A$ , you're collecting everything that is either in $A$ or in $B$ but not in $A$ . But if an element is in $B$ and also in $A$ , it's already captured by $A$ itself. So the union of $A$ and $B - A$ ends up being exactly the same as the union of $A$ and $B$ — nothing is lost, nothing extra is added.

Let's prove this formally using set properties.

Step-by-Step Proof

Start with the left-hand side.

We want to show $A \cup (B - A) = A \cup B$ .

Recall the definition: $B - A = \{ x \mid x \in B \text{ and } x \notin A \}$ .
Rewrite $B - A$ using set operations.

The set difference can be expressed as an intersection with the complement:

B - A = B \cap A^c

where $A^c$ is the complement of $A$ (relative to the universal set).

So the left side becomes:

A \cup (B \cap A^c)

Apply the distributive law. The distributive law for sets says: $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$ . Using this with $X = A$ , $Y = B$ , $Z = A^c$ :

A \cup (B \cap A^c) = (A \cup B) \cap (A \cup A^c)

Simplify $A \cup A^c$ . The union of a set and its complement is the universal set $U$ (everything under consideration):

A \cup A^c = U

So we have:

(A \cup B) \cap U

Intersection with the universal set. …

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Q.Using properties of sets, prove that for all sets AAA and BBB, A∪(B−A)=A∪BA \cup (B - A) = A \cup BA∪(B−A)=A∪B.

Why This Works

Step-by-Step Proof

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Q.Using properties of sets, prove that for all sets $A$ and $B$ , $A \cup (B - A) = A \cup B$ .

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