>>19
>(ZFCではなく)ZF上で実数は定義不可能と言いたいのですか

これに戻る
１）まず、”ZF上で実数は定義不可能”か？ について
　”実数”の意味を明確にしておく必要があるが、それを カントールの集合論における”実数”と規定する
　つまり、下記に出てくる 実数の連続性（実数の完備性 (completeness of the real numbers) とも）を、備えたものとする
２）そうすると、下記 いろいろ辿ると ”Choice principles in elementary topology and analysis Horst Herrlich”(1997)
　にたどり着いて、Equivalent are:
　"1. in R, a point x is an accumulation point of a subset A iff there exists a sequence in A＼{x} that converges to x, "
　"5. R is a Lindel¨ of space, "
　"9. the Axiom of Choice for countable collections of subsets of R."
　"Equivalent are: "
　だと。つまり、"the Axiom of Choice for countable collections of subsets of R."でも
　" in R, a point x is an accumulation point of a subset A iff there exists a sequence in A＼{x} that converges to x, "
　"R is a Lindel¨ of space, "
　までしか言えない、これが限界 （”Lindel¨ of”リンデレーエフは、下記ご参照）
３）ということは、"the Axiom of Choice for countable collections of subsets of R."を否定してしまうと
　”実数”の連続性（実数の完備性）どころか、Lindelöfさえいえない。”in R, a point x”と”iff there exists a sequence in A＼{x} that converges to x, ”
　との関係も言えない

結論：(ZFCではなく)ZF上で実数の定義では、カントールの集合論の”実数”には、到達しない
可算選択公理でさえ、R is a Lindel や in R, a point x is an accumulation point of a subset A iff there exists a sequence in A＼{x} that converges to x,
これが限界です
従属選択公理で、実数の連続性（実数の完備性）が言えるか（フルパワー選択公理でなく）

(参考)
en.wikipedia.org/wiki/Axiom_of_choice
Axiom of choice
Independence
Many theorems provable using choice are of an elegant general character: the cardinalities of any two sets are comparable

Statements implying the negation of AC
There are models of Zermelo-Fraenkel set theory in which the axiom of choice is false.
As any model of ZF¬C is also a model of ZF, it is the case that for each of the following statements, there exists a model of ZF in which that statement is true.
・There is a function f from the real numbers to the real numbers such that f is not continuous at a, but f is sequentially continuous at a, i.e., for any sequence {xn} converging to a, limn f(xn)=f(a).
・The real numbers are a countable union of countable sets.[39] This does not imply that the real numbers are countable: As pointed out above, to show that a countable union of countable sets is itself countable requires the Axiom of countable choice.
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