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Bill F
Lv 6
Bill F asked in Science & MathematicsMathematics · 1 decade ago

advanced set theory questions?

i need help with these asap. I usually contribute answers so if people can help me out here i would greatly appreciate it

Set theory

1) If g: a->b and f: g(A) -> C are injective functions, prove that f composition g is also one to one

2) Define a countable set and show that the set of natural #s is countable

3) Define an equivilance relation, and for a,b,c,d exists in Z define a/b equiv c/d if and only if ad=bc. Show that tilde is an equivalence relationship

4) what is the cardinality of the set {1/1, 1/2, 1/3 ....} .. prove it using 1 to 1 correspondence

Update:

awesome answer thanks

2 Answers

Relevance
  • Will
    Lv 5
    1 decade ago
    Favorite Answer

    1. Something like, suppose that f(g(x))=t and f(g(y))=t. Then, as f is 1-1, g(x)=g(y). Then, as g is 1-1, x=y.

    2. A Countable Set A is a set in which there exists a function f, such that f is injective, and f maps A to the naturals. The naturals are countable by the identity map.

    3) An equivilence relation is a relation which is reflexive (for all x, x~x), symmetric ( for all x,y (x~y -> y~x ) ) and transitive (for all x,y,z ( (x~y and y~z) -> x~z ) ). Clearly this relation is reflextive, because ad = da. Clearly this symmetric, because if ad=bc, then cb=da. Clearly this is transitive, because if a/b~c/d then ad=bc. And if c/d~e/f, cf=ed. Then af=be (just by the way rationals will work)

    4) This set is countable infinity. let f(a) = 1/a map the naturals to the set. This is bijective, ie. It is clearly 1-1 and onto, thus this set has the same size as N.

  • ?
    Lv 4
    4 years ago

    here's a extremely hassle-free counterexample A = {0, a million} B = {0} f(x) = 0 the place x is in A S = {0} T = {a million} S - T = {0} f(S - T) = {0} f(S) = {0} f(T) = {0} f(S) - f(T) = the empty set So f(S - T) does not equivalent f(S) - f(T) in this occasion.

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