[過去ﾛｸﾞ] 純粋・応用数学（含むガロア理論）8 (942ﾚｽ)
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202(4): 現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/05/19(水)20:58 ID:H7LP/xSH(13/15) AAS
>>188 補足
>「∈無限列 ω∋…∋1∋0 」は、上昇列であって、降下ではありませんｗ
和文しか読まない（読めない？）から、ダメなんだ
下記 Axiom of regularity で
”and that there is no infinite sequence (an) such that ai+1 is an element of ai for all i”
ですよ。分かりますかぁ〜？ ｗ
”(an) such that ai+1 is an element of ai for all i”
↓
” an ∋ ai+1 for all i ”
ですよ。分かりますかぁ〜？ｗ（＾＾；
https://en.wikipedia.org/wiki/Axiom_of_regularity
Axiom of regularity
In mathematics, the axiom of regularity (also known as the axiom of foundation) is an axiom of Zermelo?Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A. In first-order logic, the axiom reads:
∀ x,(x≠ Φ → ∃ y∈ x,(y∩ x=Φ )).
The axiom of regularity together with the axiom of pairing implies that no set is an element of itself, and that there is no infinite sequence (an) such that ai+1 is an element of ai for all i. With the axiom of dependent choice (which is a weakened form of the axiom of choice), this result can be reversed: if there are no such infinite sequences, then the axiom of regularity is true. Hence, in this context the axiom of regularity is equivalent to the sentence that there are no downward infinite membership chains.
Contents
2 The axiom of dependent choice and no infinite descending sequence of sets implies regularity
Let the non-empty set S be a counter-example to the axiom of regularity; that is, every element of S has a non-empty intersection with S.
We define a binary relation R on S by aRb:⇔ b∈ S∩ a, which is entire by assumption.
Thus, by the axiom of dependent choice, there is some sequence (an) in S satisfying anRan+1 for all n in N. As this is an infinite descending chain, we arrive at a contradiction and so, no such S exists.
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