Notes

Description: Equivalents of Form 94

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In this note we consider the equivalents of Form 94.

Brunner [1982d] proves that [94 B] implies Form 13 (lemma 6, p. 164) by showing that the negation of Form 13 implies the negation of [94 A] which implies the negation of [94 B]. Since [94 A] clearly implies [94 B], it only remains to prove that [94 A] implies Form 94 (to complete the argument for the equivalence of Form 94, [94 A] and [94 B]. Herrlich/Strecker [1997] have shown that forms Form 94 and [94 A] are equivalent. In addition they show that forms [94 A], [94 D], [94 E], [94 F], [94 G], [94 L], [94 M], [94 O] and [94 P] are equivalent.  It is clear that [94 E] implies [94 C] implies [94 D]. This answers a question of  G. Moore [1982] p 322: Does [94 D] imply [94 C]?  G. Moore also asks if Form 13 implies [94 D].  This is answered in the negative since Form 13 is true and Form 94 is false in \(\cal M6\).

It is shown in Keremedis [1998b] that [94 K] is equivalent to Form 94 and Sierpinski [1916] shows the equivalence of [94 N] and Form 94.

To get equivalence of [94 H], [94 I], and [94 J] toForm 94, we note first that (in ZF\(^0\)) every separable metric space is second countable.  From this it follows that [94 K] implies [94 H] and [94 J] implies [94 I].  Combining these with the following easy implications  gives us the desired equivalences. [94 K] \(\to\) [94 H], [94 E] \(\to\) [94 J], and [94 I] \(\to\) [94 D]. Most of the above facts were pointed out to us by K. Keremedis.

The following are definitions for [94 S]. Suppose \(x\in\Bbb R\) and \(X\subseteq \Bbb R\).
  1. \(x\) is called an accumulation point of \(X\) if every open neighborhood of \(x\) contains a point in \(X-\{x\}\).
  2. \(x\) is called a cluster point of \(X\) if every open neighborhood of \(x\) contains an infinite number of points in \(X\).
  3. \(x\) is  called a limit point of \(X\) if there is a sequence \(\{x_n: n\in\omega\}\subseteq X\) such that for every open neighborhood \(N_x\) of \(x\), there is real number \(M > 0\) such that for all \(n > M\), \(x_n\in N_x\).
In any \(T_{1}\) space, (\(\Bbb R\) with the order topology is \(T_{1}\)), a point is a cluster point if and only if it is an accumulation point. (Clearly, a cluster point is an accumulation point. Suppose \(x\) is  an accumulation point of \(X\) that is not a cluster point. Suppose \(N_x\) is a neighborhood of \(x\) that only contains a finite number of elements of \(X\), \(x_1\), \(x_2\), \(\ldots\), \(x_n\). Since \(X\) is \(T_{1}\), using induction, we can find a neighborhood \(M\) of \(x\) that does not intersect \(\{x_1, x_2,\ldots, x_n\}\). Then \(N_x\cap M\) is a neighborhood of \(x\) that does not contain any points of \(X\) different from \(x\). This contradicts the fact that \(x\) is an accumulation point.)

Howard-Rubin number: 40

Type: Equivalents

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