How does axiom of foundation prevent infinite sequence of elements?

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SUMMARY

The Axiom of Foundation, formally stated as ∀x (x ≠ ∅ → ∃y (y ∈ x ∧ y ∩ x = ∅)), prohibits the existence of infinite descending membership sequences such as x₀ ∋ x₁ ∋ x₂ ∋ ⋯. The key contradiction arises because the set x = {x₀, x₁, x₂, …} would contain elements x_{k+1} that belong both to x and to x_k, violating the condition that y ∩ x = ∅ for some y ∈ x. This establishes that no infinite ∈-chains exist, ensuring well-foundedness of sets.

PREREQUISITES

  • Formal statement and interpretation of the Axiom of Foundation (Regularity)
  • Set membership relation (∈) and set construction
  • Concept of infinite descending ∈-chains or sequences
  • Basic understanding of set theory notation and logic quantifiers

NEXT STEPS

  • Study well-founded relations and their role in set theory
  • Explore the cumulative hierarchy of sets (Vα) and its connection to foundation
  • Investigate alternative set theories without the Axiom of Foundation (e.g., non-well-founded set theory)
  • Learn formal proofs involving the Axiom of Foundation and infinite descending sequences

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Mathematics students, logicians, and researchers studying axiomatic set theory, particularly those interested in foundational aspects of sets and the prevention of pathological infinite membership chains.

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TL;DR
How does axiom of foundation prevent the infinite descending sequence of elements?
Axiom of foundation says, ##\forall x (x\neq \emptyset \to \exists y(y\in x\wedge (y\cap x=\emptyset )))##.
The book says,
"As a consequence of the Axiom of Foundation, we see that there is no
infinite descending sequence x0 ∋ x1 ∋ x2 ∋ ···, since otherwise, the
set {x0,x1,x2,...} would contradict the Axiom of Foundation."

How does it contradict the Axiom of Foundation? If y=x0, it seems to obey the axiom.
 
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Forget it. I misunderstood the statement.
 
For the sake of completeness, suppose there is such a sequence. As it says in #1, take ##x=\{x_0,x_1,\ldots\}\neq\emptyset##. Then ##x_{k+1} \in x\cap x_k## for all ##k##, contradicting the axiom.
 

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