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Countable or Uncountable…That is the question!

Countable or Uncountable…That is the question!. REVIEW. Countable 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

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Countable or Uncountable…That is the question!

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  1. Countable or Uncountable…That is the question!

  2. REVIEW • Countable • 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

  3. HOMEWORK Solutions (1) 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

  4. Are the Rational Numbers Countable? • What do we know about rational numbers? • - • - • - • - • - • -

  5. VOTINGAre the Rational Numbers countable?

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

  7. VOTINGReals in the interval (0,1) countable?

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

  9. 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, . . .

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

  11. 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

  12. 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

  13. 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

  14. How does (N x N) relate to Q+ υ {0} in size?

  15. CHALLENGE: Can you find an explicit formula for this mapping? 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) . . .

  16. 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

  17. 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}andQ- 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

  18. Hey! Q is countable! • Does this change your mind about the real numbers in the interval (0,1) being countable/uncountable?

  19. 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!!!

  20. 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 andd’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!

  21. 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!

  22. Homework • Question #1 • 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.

  23. DONE DONE DONE DONE DONE DONE DONE

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