Lecture 24 21 127 Concepts of Math 10 29 2012 Finite Cardinality Result Using Bijections Recall these two definitions A B there exists a bijection f A B A is finite n N 0 such that there exists a bijection f A n On homework you will prove that this size is unique that is a finite set has only one size that n is unique Theorem Suppose A B are disjoint finite sets Then A B A B Proof Suppose A is finite with A a and B is finite with B b and suppose A B This means there are bijections f A a and g B b WWTS there is a bijection h A B a b Define the function h A B a b by x A B h x f x if x A g x a if x B Notice that h is well defined because A B so every x A B satisfies x A or x B and certainly not both Also 1 h x a for every x A and a 1 h x a b for every x B so h x a b for every x in the domain of h We claim that the function H a b A B defined by f 1 y if 1 y a x a b H y 1 g y a if a 1 y a b is the inverse of h If this holds then we have proven that h is a bijection Notice that H is well defined because every y a b satisfies exactly one of the two inequalities given in the definition of H and furthermore f and g were given to be bijections so f 1 and g 1 are well defined functions that are bijections themselves even and even furthermore if a 1 y a b then 1 y a b so y a b the domain of g 1 Let s show that h H Id a b Let y a b be given We have two cases 1 Suppose 1 y a that is suppose y a Then h H y h H y h f 1 y f f 1 y Id a y y where we used the fact f 1 y A 2 Suppose a 1 y b that is suppose y a b Then h H y h H y h g 1 y a g g 1 y a a Id b y a a y a a y where we used the fact that g 1 y a B In either case we find h H y y and both cases are disjoint and cover all possibilities Next let s show that H h IdA B Let x A B be given We have two cases 1 1 Suppose x A Then H h x H h x H f x f 1 f x IdA x A where we used the fact that f x a 2 Suppose x B Then H h x H h x H g x a g 1 g x a a g 1 g x IdB x x where we have used the fact that g x b so a 1 g x a a b In either case we find H h x x and both cases are disjoint and cover all possibilities Thus H h 1 so h has an inverse Therefore h is a bijection Therefore A B a b a b A B More Finite Results Lemma If A and B are finite then A B A B A B You prove this You ll cover it in Recitation as well Corollary Suppose A1 A2 An are finite and pairwise disjoint remember this means any two of the sets are disjoint which is much more knowledege than just saying the intersection of all of them is the empty set Then A1 An A1 An Proof How might we prove this By induction on n the number of sets You try it We will also talk about this next week when we start combinatorics Theorem If A and B are finite then A B A B Proof By double induction on both A and B You try this We will also talk about this next week when we start combinatorics Lemma A B A B Proof Define the identity function f A B given by x A f x x Since A B this is a well defined function Note We couldn t technically define this as the usual identity function IdA because the domain and codomain might not be equal sets in essence f does the same action as IdA but has a different codomain Notice that f is injective Note It s not necessarily bijective because it might be that A 6 B Since f is injective this tells us that A B Note The result above holds for any sets not just finite ones 2 Infinite Cardinality Hilbert Hotels Let s play make believe This will help us get a handle on infinite weirdness We own a hotel There are countably infinitely many rooms numbered as Room 1 Room 2 Room 3 We want to accomodate as many people as we can to make lots of cash money and because our hotel is so swanky and accomodating our guests are totally willing to move to a new room whenever we ask them to It just takes them a couple of minutes to gather their belongings and walk down the hall to a new room We also have a loudspeaker system that allows us to communicate a message to all of the guests at once Suppose all the rooms are full It s a very busy weekend One guy knocks on the door looking for a room Can we squeeze him in If not why If so how The idea is to shift all the guests down one room and place this new guy into Room 1 The catch though is to take advantage of our loudspeaker system If we had to go and knock on everyone s door telling them to move down one room we would never actually finish we would spend all of eternity knocking on doors and delivering messages Instead we make the following announcement Attention guests If you find yourself in Room n please move to Room n 1 Thank you After five minutes the guests have all moved and Room 1 is vacant for our new guest Morally speaking we have just verified that the set N and the set N have the same cardinality for any particular object In particular say N N 0 Our rooms are still full Suppose a Scrabble convention with countably infinitely many people shows up The people are all wearing nametags with natural numbers on them so there is Person 1 Person 2 Person 3 Can we accomodate these folks How can we assign them rooms How do we move around the guests currently in the hotel The idea is to free up an infinite set of rooms and again the catch is to do this by making one blanket announcement to all …