If $A$ and $B$ are subsets of the universal set $U$, then show that (i) $A \subset A \cup B$ (ii) $A \subset B \Leftrightarrow A \cup B = B$ (iii) $(A \cap B) \subset A$

> [!TLDR] > The subset relations follow directly from the definitions of union and intersection. > (i) Every element of $A$ is in $A \cup B$, so $A \subset A \cup B$. > (ii) $A \subset B$ is equivalent to $A \cup B = B$ — each implies the other. > (iii) Every element of $A \cap B$ is in $A$, so $(A \cap B) \subset A$. --- ### Why these statements are true — the core idea Set inclusion is about **elements**. To show $X \subset Y$, you pick an arbitrary element of $X$ and argue it must belong to $Y$. The union $A \cup B$ collects everything that is in $A$ **or** in $B$ (or both). The intersection $A \cap B$ collects only what is in **both** $A$ and $B$. So: - If you are in $A$, you are certainly in $A \cup B$ — that’s (i). - If you are in $A \cap B$, you are certainly in $A$ — that’s (iii). - For (ii), the condition $A \subset B$ means every element of $A$ is already in $B$, so adding $A$ to $B$ doesn’t bring anything new; conversely, if $A \cup B = B$, then every element of $A$ is in $B$. Let’s write each proof cleanly. --- ### (i) $A \subset A \cup B$ 1. Take any element $x \in A$. 2. By definition of union, $x \in A \cup B$ if $x \in A$ **or** $x \in B$. 3. Since $x \in A$, the condition is satisfied. Hence $x \in A \cup B$. 4. Because $x$ was arbitrary, every element of $A$ is in $A \cup B$. Therefore $A \subset A \cup B$. > [!TIP] > This is the simplest subset proof: the union always contains each of its parts. No extra condition needed. --- ### (ii) $A \subset B \;\Leftrightarrow\; A \cup B = B$ We need to prove two directions. #### Direction 1: If $A \subset B$, then $A \cup B = B$ 1. **Show $A \cup B \subset B$**: Take any $x \in A \cup B$. Then $x \in A$ or $x \in B$. - If $x \in A$, then because $A \subset B$, we have $x \in B$. - If $x \in B$, trivially $x \in B$. - In either case $x \in B$. So $A \cup B \subset B$. 2. **Show $B \subset A \cup B$**: This is always true (by part (i) with $A$ and $B$ swapped, or directly: any

If A and B are subsets of the universal set U, then show that (i) A ⊂ A ∪ B (ii) A ⊂ B Leftrightarrow A ∪ B =

NCERT Exemplar · Q6

A

(i)

A \subset A \cup B

(ii)

A \subset B \Leftrightarrow A \cup B = B

(iii)

(A \cap B) \subset A

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Concept understanding — Subset Listing

Subset Listing: A First Look

Let's build this from the ground up — no jargon, just intuition first.

1. The Intuition: What does "subset" mean?

Imagine you have a set — a collection of distinct objects. For example:

Set A = {apple, banana, cherry}

Now, a subset is simply a selection of some (or all, or none) of these objects, taken from the original set.

You could pick all three → {apple, banana, cherry}
You could pick just two → {apple, banana}
You could pick just one → {cherry}
You could pick none → {} (the empty set)

Each of these is a subset of the original set.

2. The Precise Definition

Definition: A set $B$ is a subset of a set $A$ if every element of $B$ is also an element of $A$ .

We write this as:

B \subseteq A

If $B$ is not a subset of $A$ , we write:

B \not\subseteq A

Key points to remember:

Every set is a subset of itself.

Example: {apple, banana} ⊆ {apple, banana}
The empty set $\emptyset$ (or $\{\}$ ) is a subset of every set.

Why? Because it has no elements, so there's nothing to violate the condition.
If $B$ is a subset of $A$ but $B \neq A$ , we call $B$ a proper subset.

Notation: $B \subset A$ (some books use $\subsetneq$ )

3. How to "list" all subsets

Subset listing means writing down every possible subset of a given set.

Example: Set $S = \{a, b\}$

All subsets:

$\emptyset$ (empty set)
$\{a\}$
$\{b\}$
$\{a, b\}$ (the set itself)

So the list of all subsets is:

\{ \emptyset, \{a\}, \{b\}, \{a, b\} \}

How many subsets does a set have?

If a set has $n$ elements, it has exactly $2^n$ subsets.

$n = 0$ → $2^0 = 1$ subset (just the empty set)
$n = 1$ → $2^1 = 2$ subsets
$n = 2$ → $2^2 = 4$ subsets (as above)
$n = 3$ → $2^3 = 8$ subsets

Why $2^n$ ?

For each element, you have 2 choices: include it or exclude it. Multiply these choices: $2 \times 2 \times \dots \times 2$ ( $n$ times) = $2^n$ .

4. A systematic way to list subsets

For a set with $n$ elements, you can use a binary counting method:

Label each element with a position (1st, 2nd, 3rd, ...)
Count from $0$ to $2^n - 1$ in binary
Each binary number tells you which elements to include (1 = include, 0 = exclude)

Example: $S = \{a, b, c\}$ (3 elements)

Binary	Subset
000	$\emptyset$
001	$\{c\}$
010	$\{b\}$
011	$\{b, c\}$
100	$\{a\}$

Why this formula?

Okay, let's break down Subset Listing from the ground up. The core idea is simple: given a set, how do we systematically list all its subsets, and why does the formula $2^n$ work?

1. The Core Question

Imagine you have a set with $n$ elements, like $S = \{a, b, c\}$ (so $n=3$ ). A subset is any collection of elements from $S$ , including the empty set $\{\}$ and the set itself $\{a, b, c\}$ .

The key formula is:

Total number of subsets of a set with $n$ elements = $\mathbf{2^n}$

Let's see why this is true, not just memorize it.

2. The "Decision" or "Binary Choice" Reasoning

The most intuitive derivation comes from thinking about each element individually.

For each element in the original set, when building a subset, you have exactly two choices:

Include the element in the subset.
Exclude the element from the subset.

This is a fundamental, independent decision for every element.

Example with $S = \{a, b, c\}$

For element $a$ : Choose IN or OUT. (2 choices)
For element $b$ : Choose IN or OUT. (2 choices)
For element $c$ : Choose IN or OUT. (2 choices)

Since these choices are independent (choosing for $a$ doesn't affect the choice for $b$ ), the total number of distinct combinations of choices is the product of the number of choices for each element:

2 \times 2 \times 2 = 2^3 = 8

This directly gives the 8 subsets of $\{a, b, c\}$ :

$\{\}$ (all OUT)
$\{a\}$ (a IN, b OUT, c OUT)
$\{b\}$
$\{c\}$
$\{a, b\}$
$\{a, c\}$
$\{b, c\}$
$\{a, b, c\}$ (all IN)

3. The General Formula (Derivation)

For a set with $n$ elements, you have $n$ independent binary decisions. Therefore:

\text{Total subsets} = \underbrace{2 \times 2 \times \dots \times 2}_{n \text{ times}} = 2^n

This is the fundamental reason the formula holds. It's not a coincidence; it's a direct consequence of the counting principle for independent events.

4. Why This Matters for Exams

Don't just memorize $2^n$ . If a question asks "How many subsets does a set with 5 elements have?", you can instantly say $2^5 = 32$ . But if they ask why, you now have the reasoning. …

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The subset relations follow directly from the definitions of union and intersection.

Every element of $A$ is in $A \cup B$ , so $A \subset A \cup B$ .
$A \subset B$ is equivalent to $A \cup B = B$ — each implies the other.
Every element of $A \cap B$ is in $A$ , so $(A \cap B) \subset A$ .

Why these statements are true — the core idea

Set inclusion is about elements. To show $X \subset Y$ , you pick an arbitrary element of $X$ and argue it must belong to $Y$ . The union $A \cup B$ collects everything that is in $A$ or in $B$ (or both). The intersection $A \cap B$ collects only what is in both $A$ and $B$ .

So:

If you are in $A$ , you are certainly in $A \cup B$ — that’s (i).
If you are in $A \cap B$ , you are certainly in $A$ — that’s (iii).
For (ii), the condition $A \subset B$ means every element of $A$ is already in $B$ , so adding $A$ to $B$ doesn’t bring anything new; conversely, if $A \cup B = B$ , then every element of $A$ is in $B$ .

Let’s write each proof cleanly.

(i) $A \subset A \cup B$

Take any element $x \in A$ .
By definition of union, $x \in A \cup B$ if $x \in A$ or $x \in B$ .
Since $x \in A$ , the condition is satisfied. Hence $x \in A \cup B$ .
Because $x$ was arbitrary, every element of $A$ is in $A \cup B$ . Therefore $A \subset A \cup B$ .

Tip

This is the simplest subset proof: the union always contains each of its parts. No extra condition needed.

(ii) $A \subset B \;\Leftrightarrow\; A \cup B = B$

We need to prove two directions.

Direction 1: If $A \subset B$ , then $A \cup B = B$

Show $A \cup B \subset B$ : Take any $x \in A \cup B$ $x \in A \cup B$ . Then $x \in A$ $x \in A$ or $x \in B$ $x \in B$ .
- If $x \in A$ , then because $A \subset B$ , we have $x \in B$ .
- If $x \in B$ , trivially $x \in B$ .
- In either case $x \in B$ . So $A \cup B \subset B$ .
Show $B \subset A \cup B$ : This is always true (by part (i) with $A$ and $B$ swapped, or directly: any $x \in B$ is in $A \cup B$ ).
Since both inclusions hold, $A \cup B = B$ .

Direction 2: If $A \cup B = B$ , then $A \subset B$

Take any $x \in A$ .
Then $x \in A \cup B$ (by definition of union).
But $A \cup B = B$ , so $x \in B$ . …

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Q.If AAA and BBB are subsets of the universal set UUU, then show that

Why these statements are true — the core idea

(i) A⊂A∪BA \subset A \cup BA⊂A∪B

(ii) A⊂B ⇔ A∪B=BA \subset B \;\Leftrightarrow\; A \cup B = BA⊂B⇔A∪B=B

Direction 1: If A⊂BA \subset BA⊂B, then A∪B=BA \cup B = BA∪B=B

Direction 2: If A∪B=BA \cup B = BA∪B=B, then A⊂BA \subset BA⊂B

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Q.If $A$ and $B$ are subsets of the universal set $U$ , then show that

(i) $A \subset A \cup B$

(ii) $A \subset B \;\Leftrightarrow\; A \cup B = B$

Direction 1: If $A \subset B$ , then $A \cup B = B$

Direction 2: If $A \cup B = B$ , then $A \subset B$

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