Q.Write down all the subsets of the following sets

Concept: Subset enumeration using the power-set construction.

A subset of a set $S$ is any collection of elements from $S$, including the empty set $\phi$ and $S$ itself. For a set with $n$ elements, there are exactly $2^n$ subsets.

(i) $\{a\}$ has $2^1 = 2$ subsets: $\phi, \{a\}$

(ii) $\{a, b\}$ has $2^2 = 4$ subsets: $\phi, \{a\}, \{b\}, \{a,b\}$

(iii) $\{1, 2, 3\}$ has $2^3 = 8$ subsets: $\phi, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$

(iv) $\phi$ has $2^0 = 1$ subset: $\phi$ (the empty set is a subset of itself) …

Why Subsets Work This Way

When we form subsets, we're making a yes-or-no decision for each element: include it or leave it out. This binary choice for every element explains why a set with $n$ elements has $2^n$ subsets. The empty set $\phi$ is always a subset (we said "no" to everything), and the original set is always a subset of itself (we said "yes" to everything).

The key insight: subsets don't create new elements or change order—they simply select which elements to keep.

Finding All Subsets

(i) Subsets of $\{a\}$

This set has $1$ element, so we expect $2^1 = 2$ subsets.

1. The empty subset: Choose no elements → $\phi$
2. The full set: Choose the element $a$ → $\{a\}$

The subsets are: $\phi, \{a\}$

(ii) Subsets of $\{a, b\}$

This set has $2$ elements, so we expect $2^2 = 4$ subsets.

1. Choose neither element: $\phi$
2. Choose only $a$: $\{a\}$
3. Choose only $b$: $\{b\}$
4. Choose both elements: $\{a, b\}$

The subsets are: $\phi, \{a\}, \{b\}, \{a, b\}$

(iii) Subsets of $\{1, 2, 3\}$

This set has $3$ elements, so we expect $2^3 = 8$ subsets.

We can organize them by size:

The subsets are: $\phi, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}$

(iv) Subsets of $\phi$ …

Method: Listing All Subsets (Power Set) of a Finite Set

Method Name: Systematic Listing by Size

Why This Works

Every subset is formed by an independent "include or exclude" decision for each element — with $n$ elements there are $2^n$ such combinations, hence $2^n$ subsets. Listing them by size (empty set, then 1-element, then 2-element, and so on) guarantees none are missed or repeated.

Steps

Step 1: Count the elements

A set with $n$ elements has exactly $2^n$ subsets.

Step 2: Start with the empty set

$\phi$ is always a subset of every set — list it first.

Step 3: List every 1-element subset

Take each element on its own: $\{a\}$, $\{b\}$, etc.

Step 4: List every 2-element subset …

🧠 The Core Idea: Subset vs. Element

Before we list mistakes, remember the two key symbols:

• $\subset$ (subset): Every element of the first set must be in the second set.
• $\in$ (element): The entire thing on the left is a single member of the set on the right.

Mixing these up is the #1 cause of errors.

✗ Common Mistake #1: Confusing $\subset$ with $\in$

Example from the list:

Statement (v): $\{ a \} \in \{ a, b, c \}$

Why it’s wrong:

• $\{a\}$ is a set containing the letter $a$.
• $\{a, b, c\}$ contains the elements $a$, $b$, and $c$ — not the set $\{a\}$.
• So $\{a\}$ is not an element of $\{a, b, c\}$.

✓ Correct thinking:

• $\{a\} \subset \{a, b, c\}$ is true (every element of $\{a\}$ is in the big set).
• $\{a\} \in \{a, b, c\}$ is false unless the big set explicitly contains a set as an element, e.g., $\{a, \{a\}, b\}$.

How to avoid:

Ask yourself: “Is the left side a single object inside the right side, or is it a collection whose members are inside?”

✗ Common Mistake #2: Forgetting that $\subset$ requires all elements

Example from the list:

Statement (iii): $\{ 1, 2, 3 \} \subset \{ 1, 3, 5 \}$

Why it’s wrong:

• The left set has $1, 2, 3$.
• The right set has $1, 3, 5$.
• $2$ is missing from the right set. So it’s false.

✓ Correct thinking:

• For $\subset$ to be true, every element of the first set must appear in the second. One missing element = false.

How to avoid:

Check each element one by one. If even one is missing, the statement is false.

✗ Common Mistake #3: Misreading “not a subset” ($\not\subset$)

Example from the list:

Statement (i): $\{ a, b \} \not\subset \{ b, c, a \}$

Why it’s wrong:

• The left set has $a$ and $b$.
• The right set has $b, c, a$ — both $a$ and $b$ are present.
• So $\{a, b\}$ is a subset. The statement says it is not a subset — that’s false.

✓ Correct thinking:

• $\{a, b\} \subset \{b, c, a\}$ is true.
• Therefore $\{a, b\} \not\subset \{b, c, a\}$ is false.

How to avoid:

First check if it is a subset. Then apply the “not” ($\not\subset$) to decide true/false.

✗ Common Mistake #4: Overlooking the definition of the set on the right

Example from the list:

Statement (ii): $\{ a, e \} \subset \{ x : x \text{ is a vowel in the English alphabet} \}$

Why it’s correct (but often marked wrong by students):

• Vowels: $a, e, i, o, u$.
• The left set has $a$ and $e$ — both are vowels.
• So it is a subset — true.

Common error: Students sometimes think “vowel” means only $a, e, i, o, u$ but then forget to check if $a$ and $e$ are actually in that list. Or they misread the set-builder notation.

How to avoid:

Write out the actual elements of the set described in words. Then compare.

✗ Common Mistake #5: Not simplifying the set before comparing

Example from the list: …