Iterated Power Set Cardinality: Computing n(P(P(P(Φ))))n(P(P(P(\Phi))))

Iterated Power Set Cardinality: Computing n(P(P(P(Φ))))n(P(P(P(\Phi))))

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Sep 12, 2026

Power Set of the Empty Set (and cardinality intuition)

Let Φ\Phi denote the empty set, and let P()P(\cdot) denote the power set of a set. The question asks for the value of n(P(P(P(Φ)))),n\big(P(P(P(\Phi)))\big), where n(S)n(S) means the cardinality (number of elements) of the set SS.

Key facts:

  • For any finite set AA with nn elements, the cardinality of its power set is P(A)=2n|P(A)| = 2^n.
  • The empty set has 00 elements, so the power set of Φ\Phi has 20=12^0 = 1 element.
  • Iterating PP keeps reapplying the same cardinality rule: once the set has 11 element, its power set has 22 elements, etc.

We will compute this step-by-step to match the options (i)1(i)\,1, (ii)2(ii)\,2, (iii)3(iii)\,3, (iv)4(iv)\,4.

Footnotes

  1. Power set - Definition, Cardinality, Properties, Proof, Examples. (Cuemath) - States that for a set with nn elements, P(A)=2n|P(A)|=2^n, with discussion/proof ideas. 2

Step-by-step evaluation of $n(P(P(P(\Phi))))$

  1. 1
    Step 1

    Since Φ\Phi is the empty set, n(Φ)=0n(\Phi)=0.

  2. 2
    Step 2

    Use P(A)=2A|P(A)|=2^{|A|} for finite AA. Thus n(P(Φ))=2n(Φ)=20=1n(P(\Phi))=2^{n(\Phi)}=2^0=1.

    Footnotes

    1. Power set - Definition, Cardinality, Properties, Proof, Examples. (Cuemath) - States that for a set with nn elements, P(A)=2n|P(A)|=2^n, with discussion/proof ideas.

  3. 3
    Step 3

    Now the set P(Φ)P(\Phi) has 11 element, so n(P(P(Φ)))=21=2n(P(P(\Phi)))=2^{1}=2.

    Footnotes

    1. Power set - Definition, Cardinality, Properties, Proof, Examples. (Cuemath) - States that for a set with nn elements, P(A)=2n|P(A)|=2^n, with discussion/proof ideas.

  4. 4
    Step 4

    Now P(P(Φ))P(P(\Phi)) has 22 elements, so n(P(P(P(Φ))))=22=4n(P(P(P(\Phi))))=2^{2}=4.

    Footnotes

    1. Power set - Definition, Cardinality, Properties, Proof, Examples. (Cuemath) - States that for a set with nn elements, P(A)=2n|P(A)|=2^n, with discussion/proof ideas.

  5. 5
    Step 5

    Therefore the correct option is (iv)4(iv)\,4.

Power set cardinality shortcut

For finite sets, you can treat P(A)|P(A)| as 2A2^{|A|}, then iterate the exponentiation: P:01|P|: 0 \mapsto 1, then 121 \mapsto 2, then 242 \mapsto 4.

Footnotes

  1. Power set - Definition, Cardinality, Properties, Proof, Examples. (Cuemath) - States that for a set with nn elements, P(A)=2n|P(A)|=2^n, with discussion/proof ideas.

Common mistake: confusing symbols for the empty set

In many textbooks, Φ\Phi (phi) and \emptyset (empty set symbol) both mean the set with no elements. The computation above depends only on n(Φ)=0n(\Phi)=0, not on the symbol used.

Footnotes

  1. Empty Set - GeeksforGeeks - Defines empty set (symbolized by Φ\Phi or \emptyset) as a set with no elements and notes =0|\,\emptyset\,|=0 and P()P(\emptyset) has 20=12^0=1 element.

Iteration pattern (how the numbers grow)

Starting from n(Φ)=0n(\Phi)=0 and repeatedly applying P(A)=2A|P(A)|=2^{|A|}, the cardinalities evolve as: 0P1P2P4.0 \xrightarrow{P} 1 \xrightarrow{P} 2 \xrightarrow{P} 4. This corresponds exactly to n(P(Φ))=1n(P(\Phi))=1, n(P(P(Φ)))=2n(P(P(\Phi)))=2, and n(P(P(P(Φ)))))=4n(P(P(P(\Phi)))))=4.

We can summarize the three key power set results used in the iteration:

  • n(Φ)=0n(\Phi)=0 (empty set has no elements)
  • n(P(Φ))=20=1n(P(\Phi))=2^0=1.
  • n(P(P(P(Φ))))=22=4n(P(P(P(\Phi))))=2^{2}=4.

Footnotes

  1. Power set - Definition, Cardinality, Properties, Proof, Examples. (Cuemath) - States that for a set with nn elements, P(A)=2n|P(A)|=2^n, with discussion/proof ideas. 2 3

  2. Empty Set - GeeksforGeeks - Defines empty set (symbolized by Φ\Phi or \emptyset) as a set with no elements and notes =0|\,\emptyset\,|=0 and P()P(\emptyset) has 20=12^0=1 element.

Cardinality after each power-set iteration

Starting from the empty set Φ\Phi (size 0) and applying PP three times.

Why does $|P(A)|=2^{|A|}$?

Iterated power set quick checks

1 / 3
Question · Term

What is $n(\Phi)$?

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Answer · Definition

It is the empty set, so n(Φ)=0n(\Phi)=0.

Footnotes

  1. Empty Set - GeeksforGeeks - Defines empty set (symbolized by Φ\Phi or \emptyset) as a set with no elements and notes =0|\,\emptyset\,|=0 and P()P(\emptyset) has 20=12^0=1 element.

Knowledge Check

Question 1 of 4
Q1Single choice

If Φ\Phi is the empty set, what is n(P(Φ))n(P(\Phi))?