From 3bfca198afa0ab3332fc1e18bfa92ad2715fbc7a Mon Sep 17 00:00:00 2001 From: Thomas Walker Lynch Date: Wed, 12 Aug 2026 04:36:18 +0000 Subject: [PATCH] more font matter edits.. Z looking good --- document/book/TM-2026.html | 12 ++++++++---- 1 file changed, 8 insertions(+), 4 deletions(-) diff --git a/document/book/TM-2026.html b/document/book/TM-2026.html index 298595b..48e0688 100644 --- a/document/book/TM-2026.html +++ b/document/book/TM-2026.html @@ -1,4 +1,4 @@ - + @@ -214,15 +214,19 @@

- In 1908 Ernst Zermelo published an alternative system designed to avoid the known paradoxical statements of the time, even though absolute consistency remained unproven. In Zermelo's set theory, a mathematician first starts with an existing set, and then applies the Axiom of Separation using definite properties to partition out subsets Ernst Zermelo, "Untersuchungen über die Grundlagen der Mengenlehre I," Mathematische Annalen 65 (1908): 261–281.. To see how this works, consider the expression \{x \mid P(x)\}. Under unrestricted comprehension, a logician is permitted to define the predicate P(x) as x ∉ x. This produces Russell's Paradox, so the set fails to be defined. In contrast, consider the same predicate, though restricted by Zermelo's Axiom of Separation over a predefined set S, written as \dot{R} = \{x \mid x ∈ S ∧ x ∉ x\}. The only thing a person needs to know about S here is that it has already been successfully defined. So let us ask, is \dot{R} in \dot{R}? If we assume \dot{R} is a member of S, evaluating the second term forces the familiar fatal loop: if \dot{R} is in \dot{R}, it shouldn't be, and if it isn't, it should be. Thus if we assume that \dot{R} is in S, then \dot{R} can not be defined, but by definition, S is defined, and thus its members are defined. As we arrived at a contradiction, the original assumption must be false, i.e. it is wrong to assume that \dot{R} is in S. As \dot{R} is definitively not a member of S, the first term of the set comprehension rule, x ∈ S, is false, and the paradox vanishes. + In 1908 Ernst Zermelo published an alternative system designed to avoid the known paradoxical statements of the time, even though absolute consistency remained unproven. In Zermelo's set theory, a mathematician first starts with an existing set, and then applies the Axiom of Separation using definite properties to partition out subsetsErnst Zermelo, "Untersuchungen über die Grundlagen der Mengenlehre I," Mathematische Annalen 65 (1908): 261–281..

- A person might suggest defining S as the set of all definable mathematical objects, forming a universal set. However, if such a universal set S existed, the Axiom of Separation could be applied as per the proof in the prior paragraph to show \dot{R} is not in S. However, as \dot{R} is a valid, definable set, it must reside within S by the very definition of a universal set. This contradicts the premise that S contains everything. Therefore, within any system governed by the Axiom of Separation, a universal set cannot exist. + To see how this works, begin with the unrestricted case. A logician is permitted to define the predicate P(x) as x ∉ x. Then \{x \mid P(x)\} produces Russell's paradox, so the set fails to be defined. This is inconsistent with the founding assumption that any predicate would work, so it is a problem. In contrast, consider the same condition, though restricted by Zermelo's Axiom of Separation over a predefined set S, written as \dot{R} = \{x \mid x ∈ S ∧ x ∉ x\}. Now the predicate has two terms. The only thing a person needs to know about S here is that it has already been successfully defined; we don't need to know what that definition is. Now assume \dot{R} is in S. That gives the second term of the condition authority, which enables the familiar fatal loop: if \dot{R} is in \dot{R}, it shouldn't be, and if it isn't, it should be. And thus it is clear that the initial assumption, that \dot{R} is in S, must be wrong, and \dot{R} is not in S. Authority returns to the first term, x ∈ S, which is false, so the condition is false. No contradiction follows, and the paradox vanishes.

- The authority to remove Russell's Paradox set formulation comes from the set S. If we know its definition, then the authority comes through that definition. However, if we merely stipulate that S must be defined, then we are expressing our authority through S by declaring, "Undefined sets are not allowed." In the explanation above, it is only after discovering a set is undefined that we conclude it is not a member of S. I sometimes wonder how mathematics might have evolved had Frege simply taken that approach. We take this question up again in chapter , Computational Naturalism, and discover there is a deeper issue. + A person might suggest defining S as the set of all definable mathematical objects, forming a universal set that can be used in any set formulation by this method. However, if such a universal set S existed, the Axiom of Separation could be applied as per the proof in the prior paragraph to show \dot{R} is not in S. That contradicts the premise that S contains everything. Therefore, within any system governed by the Axiom of Separation, a universal set cannot existibid. +

+ +

+ The authority to remove Russell's Paradox set formulation comes from the set S. If we know its definition, then the authority comes through that definition. However, if we merely stipulate that S must be defined, then we are expressing our authority through S by declaring, "Undefined sets are not allowed." In the explanation above, it is only after discovering a set is undefined that we conclude it is not a member of Sibid. I sometimes wonder how mathematics might have evolved had Frege simply taken that approach. We take this question up again in chapter , Computational Naturalism, and discover there is a deeper issue.

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