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Failed to connect to MySQL: (1203) User howardkn already has more than 'max_user_connections' active connectionsPath: ...!Xl.tags.giganews.com!local-4.nntp.ord.giganews.com!news.giganews.com.POSTED!not-for-mail NNTP-Posting-Date: Sun, 18 Aug 2024 14:17:18 +0000 Subject: Re: Replacement of Cardinality (infinite middle) Newsgroups: sci.logic,sci.math References: <5f795e1a-346b-43f7-a2d2-7844591f5296@att.net> <-oGdnWXm-ZVn1iT7nZ2dnZfqn_ednZ2d@giganews.com> <11887364-602b-4496-8f37-aa6ec7d9f69c@att.net> <2ce53910-5bb0-4ebd-805b-dccc0b21dc13@att.net> <30967b25-6a7e-4a67-a45a-99f5f2107b74@att.net> <58c50fcb-41ea-4ac3-9791-81dafd4b7a59@att.net> <29fc2200-8ddc-43fe-9130-ea49301d3c5d@att.net> From: Ross Finlayson Date: Sun, 18 Aug 2024 07:17:06 -0700 User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:38.0) Gecko/20100101 Thunderbird/38.6.0 MIME-Version: 1.0 In-Reply-To: <29fc2200-8ddc-43fe-9130-ea49301d3c5d@att.net> Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Message-ID: Lines: 196 X-Usenet-Provider: http://www.giganews.com X-Trace: sv3-oCVRkSKFNKgxrLFDB5hLSRXpw/nDrQSqfCE7lD8w0Pvzy4Zx9UbTuZM1nY9QyIpZvqCOmdibBPdRu+9!mkGt/gJiRyUDt5JLx86yrHLfC70rt+JoGIaRcUFL8xYV5e3Qe49PusN0+sokGzz1BPBUVdxUI0c= X-Complaints-To: abuse@giganews.com X-DMCA-Notifications: http://www.giganews.com/info/dmca.html X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.40 Bytes: 8262 On 08/17/2024 02:12 PM, Jim Burns wrote: > On 8/16/2024 10:28 PM, Ross Finlayson wrote: >> On 08/13/2024 08:37 PM, Jim Burns wrote: >>> On 8/13/2024 9:03 PM, Ross Finlayson wrote: >>>> On 08/12/2024 09:25 PM, Jim Burns wrote: >>>>> On 8/12/2024 8:28 PM, Ross Finlayson wrote: > >>>>>> It's like yesterday, >>>>>> in this thread with the subject of it >>>>>> talking about >>>>>> "infinite in the middle and >>>>>> always with both ends", > >>>>> "ALWAYS with both ends" is finite. >>> >>>> If it's infinite in the middle >>> >>> If >>> it's infinite in the middle and >>> its non.{} subsets always have both ends, >>> then >>> it's not infinite in the middle. > >> So, you seem to imply that >> the integers by your definition, > > Paul Gustav Samuel Stäckelᵖᵍˢˢ > (20 August 1862, Berlin – 12 December 1919, Heidelberg) > >> the integers by your definition, >> by simply assigning 1 and -1 to the beginning, >> then interleaving them, >> and filling in as like a Pascal's Triangle of sorts, >> or pyramidal numbers, that >> that's, not, infinite? >> >> Or, the rationals in the usual assignment of >> assigning them integer values >> and all the criss-crossing and >> from either end, building in the middle, >> not, infinite? > > ℕ ℤ ℚ and ℝ are each infiniteᵖᵍˢˢ, > each not.finiteᵖᵍˢˢ, > in the sense of Stäckel's finiteᵖᵍˢˢ, > by lemma 1. > > Lemma 1. > ⎛ No set B has both > ⎝ finiteᵖᵍˢˢ order ⟨B,<⟩ and infiniteᵖᵍˢˢ order ⟨B,⩹⟩. > > Definition. > ⎛ An order ⟨B,<⟩ of B is finiteᵖᵍˢˢ iff > ⎜ each non.empty subset S ⊆ B holds > ⎝ both min[<].S and max[<].S > > A finiteᵖᵍˢˢ set has a finiteᵖᵍˢˢ order. > An infiniteᵖᵍˢˢ set doesn't have a finiteᵖᵍˢˢ order. > > ℕ ℤ ℚ and ℝ each have infiniteᵖᵍˢˢ orders. > In the standard order, > ℕ ℤ ℚ and ℝ are subsets of ℕ ℤ ℚ and ℝ with > 0 or 1 ends. > Thus, the standard order is infiniteᵖᵍˢˢ. > Thus, by lemma 1, no non.standard order is finiteᵖᵍˢˢ. > > They do not have any finiteᵖᵍˢˢ order. > Whatever non.standard order you propose, > you are proposing an infiniteᵖᵍˢˢ order; > you are proposing an order with > some _subset_ with 0 or 1 ends. > > One more time: > In a finiteᵖᵍˢˢ order, > _each non.empty subset_ is 2.ended. > Two ends for the set as a whole isn't enough > to make the order finiteᵖᵍˢˢ. > > ---- > Lemma 1. > No set B has both > finiteᵖᵍˢˢ order ⟨B,<⟩ and infiniteᵖᵍˢˢ order ⟨B,⩹⟩. > > ⎛ Assume otherwise. > ⎜ Assume finiteᵖᵍˢˢ ⟨B,<⟩ and infiniteᵖᵍˢˢ ⟨B,⩹⟩. > ⎜ > ⎜ When ordered by '<', > ⎜ there is a first initial.segment[<] ⟨x₀,xⱼ⟩ᑉ > ⎜ such that, when ordered by '⩹', > ⎜ ⟨⟨x₀,xⱼ⟩ᑉ,⩹⟩ is infiniteᵖᵍˢˢ > ⎜ and > ⎜ ⟨⟨x₀,xⱼ₋₁⟩ᑉ,⩹⟩ is finiteᵖᵍˢˢ > ⎜ > ⎜ However, > ⎜ ⟨x₀,xⱼ⟩ᑉ = ⟨x₀,xⱼ₋₁⟩ᑉ∪{xⱼ} > ⎜ By lemma 2, > ⎜ there is no set B, no element x, no order '⩹' > ⎜ such that > ⎜ B is finiteᵖᵍˢˢ and B∪{x} is infiniteᵖᵍˢˢ. > ⎝ Contradiction. > > Therefore, > no set B has both > finiteᵖᵍˢˢ order ⟨B,<⟩ and infiniteᵖᵍˢˢ order ⟨B,⩹⟩. > > ---- > Lemma 2. > There is no set B, no element x, no order '⩹' > such that > ⟨B,⩹⟩ is finiteᵖᵍˢˢ and ⟨B∪{x},⩹⟩ is infiniteᵖᵍˢˢ. > > ⎛ Assume otherwise. > ⎜ Assume ⟨B,⩹⟩ is finiteᵖᵍˢˢ > ⎜ and ⟨B∪{x},⩹⟩ is infiniteᵖᵍˢˢ > ⎜ > ⎜ Consider non.empty S ⊆ B∪{x} > ⎜ Either a) b) c) or d) is satisfied, and, > ⎜ in each case, S is 2.ended[⩹] > ⎜ > ⎜ a) > ⎜ x ∉ S > ⎜ S ⊆ B > ⎜ S ⊆ B∪{x} is 2.ended[⩹] > ⎜ > ⎜ b) > ⎜ x ∈ S > ⎜ x = min[⩹].S > ⎜ x ≠ max[⩹].S = max[⩹].(S\{x}) > ⎜ S ⊆ B∪{x} is 2.ended[⩹] > ⎜ > ⎜ c) > ⎜ x ∈ S > ⎜ x ≠ min[⩹].S = min[⩹].(S\{x}) > ⎜ x = max[⩹].S > ⎜ S ⊆ B∪{x} is 2.ended[⩹] > ⎜ > ⎜ d) > ⎜ x ∈ S > ⎜ x ≠ min[⩹].S = min[⩹].(S\{x}) > ⎜ x ≠ max[⩹].S = max[⩹].(S\{x}) > ⎜ S ⊆ B∪{x} is 2.ended[⩹] > ⎜ > ⎜ Each non.empty S ⊆ Bu{x} is 2.ended[⩹] > ⎜ ⟨B∪{x},⩹⟩ is finiteᵖᵍˢˢ > ⎝ Contradiction. > > Therefore, > there is no set B, no element x, no order '⩹' > such that > ⟨B,⩹⟩ is finiteᵖᵍˢˢ and ⟨B∪{x},⩹⟩ is infiniteᵖᵍˢˢ. > > So, with "infinite in the middle", it's just that the natural order 0, infinity - 0, 1, infinity - 1, ========== REMAINDER OF ARTICLE TRUNCATED ==========