Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
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
awesome answer thanks
2 Answers
- WillLv 51 decade agoFavorite 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 44 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.