Let $A = \phi$. Find $P(P(A))$.

> [!TLDR] Applying the power-set operation twice to the empty set gives $P(P(\phi))=\{\phi,\{\phi\}\}$, a set with 2 elements. > [!FORMULA] The power set $P(S)$ is the set of all subsets of $S$, and $n(P(S))=2^{n(S)}$. > > 1. $A=\phi$, so $n(A)=0$. The only subset of $\phi$ is $\phi$ itself, so $P(A)=\{\phi\}$. Check via the formula: $n(P(A))=2^0=1$ ✓. > > 2. Now find $P(P(A))=P(\{\phi\})$. The set $\{\phi\}$ has exactly one element (namely $\phi$), so $n(\{\phi\})=1$. > …

Let A = φ. Find P(P(A)).

Exercise 3.2 · Q6

A = \phi

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Concept understanding — Representation of Sets

Representation of Sets

The Intuition

Imagine you have a bag of marbles. The bag itself is the set, and each marble inside is an element of that set. The only thing that matters is which marbles are inside — not the order you list them, not how many times you name each marble, just whether a marble belongs to the bag or not.

A set is simply a well-defined collection of distinct objects. "Well-defined" means there is no ambiguity: for any object you consider, you can say clearly "yes, it belongs" or "no, it does not."

Now, how do you describe which marbles are in the bag? There are three standard ways to do this in mathematics.

1. Roster Form (Tabular Form)

You literally list all the elements, separated by commas, and enclose them in curly braces $\{ \}$ .

Example: The set of vowels in the English alphabet:

V = \{a, e, i, o, u\}

Example: The set of natural numbers less than 5:

A = \{1, 2, 3, 4\}

Watch out

In roster form, order does not matter. $\{1, 2, 3\}$ and $\{3, 1, 2\}$ are the same set. Also, repeating an element is meaningless — $\{1, 1, 2\}$ is just $\{1, 2\}$ .

When the set has too many elements to list, you can use an ellipsis ( $\dots$ ) if the pattern is clear.

Example: The set of all natural numbers:

\mathbb{N} = \{1, 2, 3, 4, \dots\}

Example: The set of even numbers between 2 and 20:

E = \{2, 4, 6, 8, \dots, 20\}

2. Set-Builder Form (Rule Form)

Instead of listing elements, you describe the property that all elements share. The general structure is:

\{ x \mid \text{condition(s) that } x \text{ must satisfy} \}

The vertical bar $\mid$ is read as "such that." The variable $x$ is a placeholder for any element.

Example: The set of vowels from earlier:

V = \{ x \mid x \text{ is a vowel in the English alphabet} \}

Example: The set of natural numbers less than 5:

A = \{ x \mid x \in \mathbb{N}, \; x < 5 \}

Here $\in$ means "belongs to" or "is an element of." So this reads: "the set of all $x$ such that $x$ is a natural number and $x$ is less than 5."

Tip

Set-builder form is powerful when the roster form would be impossibly long or when the pattern isn't obvious. For example, the set of all real numbers between 0 and 1:

$\{ x \mid x \in \mathbb{R}, \; 0 < x < 1 \}$

You cannot list these — there are infinitely many.

3. Venn Diagram Representation

This is a visual method. You draw a closed curve (usually a circle or oval) to represent the set, and place the elements inside it. The rectangle around everything represents the universal set (all objects under consideration).

Example: The set $A = \{1, 2, 3\}$ inside the universal set $U = \{1, 2, 3, 4, 5\}$ :

  ┌─────────────┐
  │   U         │
  │  ┌─────┐    │
  │  │ 1   │    │
  │  │ 2 3 │    │
  │  └─────┘    │
  │     4 5     │
  └─────────────┘

Venn diagrams are especially useful when you have two or more sets and want to see their relationships (union, intersection, difference).

The Precise Statement …

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Applying the power-set operation twice to the empty set gives $P(P(\phi))=\{\phi,\{\phi\}\}$ , a set with 2 elements.

[!FORMULA] The power set $P(S)$ is the set of all subsets of $S$ , and $n(P(S))=2^{n(S)}$ .

$A=\phi$ , so $n(A)=0$ . The only subset of $\phi$ is $\phi$ itself, so $P(A)=\{\phi\}$ . Check via the formula: $n(P(A))=2^0=1$ ✓.
Now find $P(P(A))=P(\{\phi\})$ . The set $\{\phi\}$ has exactly one element (namely $\phi$ ), so $n(\{\phi\})=1$ .

…

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Q.Let A=ϕA = \phiA=ϕ. Find P(P(A))P(P(A))P(P(A)).

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Q.Let $A = \phi$ . Find $P(P(A))$ .

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