3.12ABBC030109

3.12ABBC030109 - 1 3.12 ABBC Recall the following reduced...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 3.12. ABBC. Recall the following reduced table for AB from section 3.5. REDUCED AB 1. A ∪ . fA ⊆ B ∪ . gA. INF. AL. ALF. FIN. NON. 2. A ∪ . fA ⊆ B ∪ . gB. INF. AL. ALF. FIN. NON. 3. A ∪ . fA ⊆ B ∪ . gC. INF. AL. ALF. FIN. NON. 4. C ∪ . fA ⊆ B ∪ . gA. INF. AL. ALF. FIN. NON. 5. C ∪ . fA ⊆ B ∪ . gB. INF. AL. ALF. FIN. NON. 6. C ∪ . fA ⊆ B ∪ . gC. INF. AL. ALF. FIN. NON. The reduced table for BC is obtained from the reduced table for AB via the permutation that sends A to B, B to C, and C to A. We use 1'-6' to avoid confusion. REDUCED BC 1’. B ∪ . fB ⊆ C ∪ . gB. INF. AL. ALF. FIN. NON. 2’. B ∪ . fB ⊆ C ∪ . gC. INF. AL. ALF. FIN. NON. 3’. B ∪ . fB ⊆ C ∪ . gA. INF. AL. ALF. FIN. NON. 4’. A ∪ . fB ⊆ C ∪ . gB. INF. AL. ALF. FIN. NON. 5’. A ∪ . fB ⊆ C ∪ . gC. INF. AL. ALF. FIN. NON. 6’. A ∪ . fB ⊆ C ∪ . gA. INF. AL. ALF. FIN. NON. This results in 36 ordered pairs, which we divide into six cases. We begin with two Lemmas. We will determine the status of all attributes INF, AL, ALF, FIN, NON, for all ordered pairs. LEMMA 3.12.1. C ∪ . fX ⊆ B ∪ . gY, Z ∪ . fB ⊆ C ∪ . gW has ¬ INF, ¬ FIN. Proof: Let f be as given by Lemma 3.2.1. Let g ∈ ELG be given by g(n) = 2n+1. Let C ∪ . fX ⊆ B ∪ . gY, Z ∪ . fB ⊆ C ∪ . gW, where A,B,C are nonempty. Clearly fB ∩ 2N ⊆ C. By C ⊆ B ∪ gY, we have fB ∩ 2N ⊆ B. Hence by Lemma 3.2.1, fB is cofinite. Hence B is infinite. This establishes that ¬ FIN. Also Z is finite. This establishes that ¬ INF. QED LEMMA 3.12.2. C ∪ . fX ⊆ B ∪ . gY, Z ∪ . fB ⊆ C ∪ . gW, B ∩ fB = ∅ has ¬ NON. 2 Proof: We can continue the proof of Lemma 3.12.1. Using fB is cofinite and B is finite, we obtain an immediate contradiction from B ∩ fB = ∅ . QED We use Lemmas 3.12.1 and 3.12.2 in cases 5,6 below. part 1. A ∪ . fA ⊆ B ∪ . gA. 1,1’. A ∪ . fA ⊆ B ∪ . gA, B ∪ . fB ⊆ C ∪ . gB. ¬ INF. ¬ AL. ¬ ALF. ¬ FIN. ¬ NON. 1,2’. A ∪ . fA ⊆ B ∪ . gA, B ∪ . fB ⊆ C ∪ . gC. ¬ INF. ¬ AL. ¬ ALF. ¬ FIN. ¬ NON. 1,3’. A ∪ . fA ⊆ B ∪ . gA, B ∪ . fB ⊆ C ∪ . gA. ¬ INF. ¬ AL. ¬ ALF. ¬ FIN. ¬ NON. 1,4’. A ∪ . fA ⊆ B ∪ . gA, A ∪ . fB ⊆ C ∪ . gB. INF. AL. ALF. FIN. NON. 1,5’. A ∪ . fA ⊆ B ∪ . gA, A ∪ . fB ⊆ C ∪ . gC. INF. AL. ALF. FIN. NON. 1,6’. A ∪ . fA ⊆ B ∪ . gA, A ∪ . fB ⊆ C ∪ . gA. INF. AL. ALF. FIN. NON. The following pertains to 1,4’, 1,6’. LEMMA 3.12.3. A ∪ . fA ⊆ B ∪ . gA, A ∪ . fB ⊆ C ∪ . gX has INF, ALF provided X ∈ {A,B}, even for EVSD. Proof: Let f,g ∈ EVSD. Let n be sufficiently large. By Lemma 3.2.5, let A ⊆ [n, ∞ ) be infinite, where A is disjoint from f(A ∪ fA) ∪ g(A ∪ fA). Let B = (A ∪ fA)\gA, and C = (A ∪ fB)\gX....
View Full Document

This note was uploaded on 08/05/2011 for the course MATH 366 taught by Professor Joshua during the Fall '08 term at Ohio State.

Page1 / 14

3.12ABBC030109 - 1 3.12 ABBC Recall the following reduced...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online