NOTES WEEK 02 DAY 1. THEOREM 0.3. Let A, B and C be sets. Then

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1 NOTES WEEK 02 DAY 1 SCOT ADAMS LEMMA P, Q, R, rp or pq&rqs rp p or Qq&pP or Rqs. THEOREM 0.2. Let A and B be sets. Then (i) A X B B X A, and (ii) A Y B B Y A THEOREM 0.3. Let A, B and C be sets. Then (i) pa X Bq X C A X pb X Cq, (ii) pa Y Bq Y C A Y pb Y Cq, (iii) A X pb Y Cq pa X Bq Y pa X Cq, (iv) A Y pb X Cq pa Y Bq X pa Y Cq, and and Proof. We leave (i) and (ii) as unassigned homework. HW#5. Proof of (iv): We wish to show: Part (iii) x P A Y pb X Cqq px P pa Y Bq X pa Y Cqq. Given x. We wish to show: px P AYpBXCqq px P paybqxpaycqq. We have rx P A Y pb X Cqs rpx P Aq or px P B X Cqs rpx P Aq or ppx P Bq&px P Cqqs rppx P Aq or px P Bqq&ppx P Aq or px P Cqqs rpx P A Y Bq&px P A Y Cqs rx P pa Y Bq X pa Y Cqs. End of proof of (iv). Date: September 13, 2016 Printout date: November 29,

2 2 SCOT ADAMS The next lemma says that & distributes over &. In fact, it s useful to know that each of & and or distributes over each of & and or, four separate resuls in propositional calculus. LEMMA 0.4. For any propositions P, Q and R, we have: P & pq&rq pp &Qq & pp &Rq. THEOREM 0.5. Let A, B and C be sets. Then AzpB Y Cq pazbq X pazcq. Proof. We wish to show, for all x, that x P AzpB Y Cq x P pazbq X pazcq. Let x be given. We wish to prove: x P AzpB Y Cq x P pazbq X pazcq. We have x P AzpB Y Cq rx P As & rnotpx P pb Y Cqqs rx P As & rnotppx P Bq or px P Cqqs rx P As & rpnotpx P Bqq&pnotpx P Cqqs prx P As&rpnotpx P Bqqsq & prx P As&rpnotpx P Cqqsq px P AzBq & px P AzCq x P pazbq X pazcq, as desired. Assigned HW#6. THEOREM 0.6. For any set A, we have H Ď A. Proof. Let a set A be given. We wish to show that H Ď A. We P H, x P A. Then H Ď A, as desired. DEFINITION 0.7. For any set S, we define $ & a, if S tau; ELTpSq :, if S H %, if S has more than one element.

3 NOTES WEEK 02 DAY 1 3 Then ELTt7u 7, ELTpHq, ELTt1, 2, 3u. Also we have: ELTtt1, 2, 3uu t1, 2, 3u, ELTtHu H. Sets are unordered. For example: FACT 0.8. For all a, b, x, y, we have: j j ta, bu tx, yu rpa xq&pb yqs or rpa yq&pb xqs. NOTE TO SELF: Next year, insist that any construction involving is equal to. So modify Definition 0.9 below so that, for any a, we have p, aq pa, q. NOTE TO SELF: Let x, y P R. Later on, we ll define R 2 R t1,2u. To distinguish between px, yq P R 2 R t1,2u and px, yq P R ˆ R, let s use px, yq P R 2 and xxx, yyy P R ˆ R. So, in Definition 0.9 below, let s use xxx, yyy instead of px, yq. DEFINITION 0.9. For any x, y, we define # tx, tx, yuu, if x y; px, yq : txu, if x y. The notation px, yq is read the ordered pair x, y. We have p1, 2q t1, t1, 2uu t2, t2, 1uu p2, 1q. Also, p3, 3q t3u. Also, p1, t2, 5uq t1, t1, t2, 5uuu. NOTE TO SELF: Next year, Fact 0.10 below will have to be modified to insist that is not equal to a, b, x or y. Perhaps it needs to be a convention means for every object p not equal to. Ordered pairs are ordered: FACT For all a, b, x, y, we have: rpa, bq px, yqs rpa xq&pb yqs. DEFINITION For any sets X and Y, we define X ˆ Y : tpa, bq a P X, b P Y u. Then t1, 2u ˆ t2, 3, 4u tp1, 2q, p1, 3q, p1, 4q, p2, 2q, p2, 3q, p2, 4qu t t1, t1, 2uu, t1, t1, 3uu, t1, t1, 4uu, t2u, t2, t2, 3uu, t2, t2, 4uu u.

4 4 SCOT ADAMS We showed how to visualize X ˆ Y in a Venn diagram. DEFINITION Let X, Y, f be sets. Then f : X Y means both f Ď X ˆ Y P c P Y, p rpa, bq, pa, cq P fs ñ rb cs q. N.B.: The P c P Y, p rpa, bq, pa, cq P fs ñ rb cs q is sometimes called the vertical line test. We did some an example tpx, yq P RˆR xy 1u and a non-example tpx, yq P R ˆ R x 2 ` y 2 1u. The notation f : X Y is read f is a partial function from X into Y or f is a function from a subset of X onto a subset of Y or f is a function from a part of X to a part of Y. Sometimes to a part of Y is shortened to into Y. Sometimes into Y is shortened to to Y. We showed two different ways to visualize a partial function. DEFINITION Let X and Y be sets. Let f : X Y. Then domrfs : ta P X Db P Y s.t. pa, bq P fu imrfs : tb P Y Da P X s.t. pa, bq P fu. These are called the domain and image of f, respectively. We showed two different ways to visualize the domain of a function. We also showed two different ways to visualize the image of a function. (First, with X ˆ Y as a rectangle and f inside that rectangle. Second, with f projecting points downward from a two-dimensional X to a horizontal line segment Y positioned below part of X.) DEFINITION Let X and Y be sets. Let f : X Y. Then, for all a, we define fpaq : ELTtb P Y pa, bq P fu. Let f : tp1, 2q, p2, 3q, p3, 5qu : t1, 2, 3, 4u t2, 3, 4, 5u. Then fp1q 2 and fp2q 3 and fp3q 5 and fp4q and fp100q. We defined f : X Ñ Y to mean: f : X Y and domrfs X. We defined f : X ą Y to mean: f : X Y and imrfs Y. We defined f : X Ñą Y to mean: f : X Y and domrfs X and imrfs Y.

5 NOTES WEEK 02 DAY 1 5 DEFINITION Let X and Y be sets. Let f Ď X ˆ Y. Then f : X Ă Y means both f : X Y b P domrfs, rpfpaq fpbqq ñ pa bqs. We defined f : X ĂÑ Y to mean: f : X Ă Y and domrfs X. We defined f : X Ă ą Y to mean: f : X Ă Y and imrfs Y. We defined f : X ĂÑą Y to mean: f : X Ă Y and domrfs X and imrfs Y. A partial function f : X Y is said to be injective or 1-1 if f : X Ă Y. The notation 1-1 is read one-to-one. A partial function f : X Y is said to be surjective onto Y or onto Y if f : X ą Y. A partial function f : X Y is said to be bijective from X onto Y if f : X Ă ą Y. We showed two ways to visualize injections. We showed two ways to visualize surjections. We showed two ways to visualize bijections.

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