Countable or Uncountable…That is the question!

REVIEW Countable Countably Infinite Uncountable Empty set, finite set or countably infinite Countably Infinite The set is a non-empty, non-finite set, and there exists a bijection between N and the set. Uncountable Not countable

HOMEWORK Solutions (1) Yes (2) Yes The function f(n) = 2n is the desired bijection. (2) Yes The function f(n) = is the desired bijection. -n/2 if n is even (n-1)/2 if n is odd

Are the Rational Numbers Countable? What do we know about rational numbers? - -Dense -N in Z in Q -Def of rational #

VOTING Are the Rational Numbers countable? (A) YES (B) NO (C) UNSURE

What about the interval (0,1) What do we know about this interval? -

VOTING Reals in the interval (0,1) countable? (A) YES (B) NO (C) UNSURE

Let’s prove some things to attack these questions! If A & B are disjoint countably infinite sets then AυB is countable.

Proof Since A is countable there exists a bijection f : N  A such that f (i) = ai Since B is countable there exists a bijection g : N  B such that g (i) = bi Construct a function h (i) that orders the elements of A and B in the following way: a1, b1, a2, b2, a3, b3, . . .

Our function h (i) h (i) = Why is h a bijection? RECORD these observations on your worksheet

Let’s prove some things to attack these questions! If A & B are disjoint countably infinite sets then AυB is countable. If A is a countably infinite set and B is a subset of A then B is countable. DONE

If A is a countably infinite set and B is a subset of A then B is countable. Case I: If B is the empty set or a finite set then B is countable. Case II: B is an infinite set Since A is countable we can write the elements of A in the order a1, a2, a3, . . . If B is a subset of A then an infinite number of elements in the above sequence are elements of B. Thus the elements of B form a subsequence (c1, c2, c3,. . .) of the sequence a1, a2, a3, . . ., thus we may order the elements of B as b1, b2, b3, . . . where bk = ck and the function f (i) = bi is a bijection between N and B

Let’s prove some things to attack these questions! If A & B are disjoint countably infinite sets then AυB is countable. If A is a countably infinite set and B is a subset of A then B is countable. DONE DONE

How does (N x N) relate to Q+ υ {0} in size? . 0,2 1,2 2,2 3,2 . . . 0,1 1,1 2,1 3,1 0,0 1,0 2,0 3,0

Can you find an explicit formula for this mapping? CHALLENGE: . 0,2 1,2 2,2 3,2 . . . 0,1 1,1 2,1 3,1 0,0 1,0 2,0 3,0 f(1) = (0,0) f(2) = (1,0) f(3) = (0,1) f(4) = (0,2) f(5) = (1,1) f(6) = (2,0) f(7) = (3,0) f(8) = (2,1) f(9) = (1,2) .

Prove It! Now that we know that NxN is countable we can show that Q is countable. Use the facts we have deduced to show that Q is countable

Proof that Q is countable We know that Q+ υ {0} can be thought of as a subset of NxN Similarly Q- can be thought of as a subset of NxN Q+ υ {0} and Q- are countable because they are subsets of a countable set. We have shown that the union of two countable sets is also countable so (Q+ υ {0}) υ Q- = Q is countable

Hey! Q is countable! Does this change your mind about the real numbers in the interval (0,1) being countable/uncountable? (A) YES (B) NO (C) UNSURE

Cantor’s Diagonalization Argument Cantor proved that the interval of real numbers (0,1) is… We start by noting that each real number in the interval (0,1) has a unique decimal representation of the form 0.d1d2d3d4… (where each di is a number from 0-9). And where decimals with period 1 cannot repeat with the number 9. UNCOUNTABLE!!!

Proof (by contradiction) Assume that f is a bijection from N  (0,1). Then we may say: f(1) = A = 0.a1a2a3a4… f(2) = B = 0.b1b2b3b4… (Where A,B,C,D are distinct real numbers in (0,1)) f(3) = C = 0.c1c2c3c4… f(4) = D = 0.d1d2d3d4… . Choose a digit from 0 to 8 and we will call this a’1 such that a’1 ≠a1 Similarly choose a digit from 0 to 8 for b’1 such that b’2≠ b2 Similarly choose c’3≠ c3 and d’4≠ d4 . . . Clearly f does not map a natural number to the real number 0.a’1b’2c’3d’4… So f is not a bijection. Contradiction!

This proof show us that the interval (0,1) is actually a larger size of infinity than the natural numbers. HENCE (0,1) is a larger size of infinity than Q as well!

Homework Question #1 Question #2 Today we identified two kinds of infinity: the size of the natural numbers and the size of the interval (0,1). Show that there is a bijection between the interval (0,1) and the set of real numbers. What does the existence of this bijection imply about these two sets? Question #2 Can you find a different size of infinity? That is a set that cannot be put into a bijection with N or the interval (0,1). To help you with this problem research the findings of Paul Cohen.

DONE DONE DONE DONE DONE DONE DONE