Let A=\∅,\∅\\ . How many elements does P(A) have, and what is correct about A ∈ A under regularity?

4 elements, and A∈ A is impossible

3 elements, and A∈ A is impossible

2 elements, and A∈ A is possible

4 elements, and A∈ A is possible

Exercise · ZW-ZERMELO-FRAENKEL-AXIOMS-009 · ★★★ Advanced

Let \( A=\{\varnothing,\{\varnothing\}\} \). How many elements does \( \mathcal{P}(A) \) have, and what is correct about \( A \in A \) under regularity?

Choose one option.

Show answer

A. 4 elements, and \( A\in A \) is impossible

Explanation

Since \( |A|=2 \), \( |\mathcal{P}(A)|=2^2=4 \). Regularity excludes \( A\in A \). Counting only nonempty subsets gives a wrong result.

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Review the concept: Zermelo–Fraenkel axioms

The Zermelo-Fraenkel axioms (ZF) are the standard axiom system of set theory, formulated in first-order logic. They allow the usual mathematics to be built from the primitive notions of set and membership (\( \in \)).

\[\forall x\,\forall y\,(\forall z\,(z\in x \leftrightarrow z\in y)\rightarrow x=y)\]

Let \( A=\{\varnothing,\{\varnothing\}\} \). How many elements does \( \mathcal{P}(A) \) have, and what is correct about \( A \in A \) under regularity?

Explanation

Review the concept: Zermelo–Fraenkel axioms

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