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Stumped on 2 easy function problems!

  1. May 3, 2008 #1
    The first question I have is simple, but when I attempted it, I got stuck.

    I'm trying to prove that if f:X->Y and A & B are subsets of X, that f(A intersect B) is a subset of f(A) intersect f(B).

    I started by trying to show set containment, beginning with an arbitrary element in f(A intersect B). However, I cannot figure out how to transition into the right hand side of the problem.

    ----------------------------

    The second question I have is proving that if A and B are finite sets having the same cardinality and f:A->B is one-to-one then f is onto.

    I missed class this day and can't figure out what cardinality is by reading the chapter.

    Someone please help! =\
     
  2. jcsd
  3. May 3, 2008 #2
    1. If [tex]y \in f(A \cap B)[/tex], then [tex]\exists x \in A \cap B[/tex] such that f(x) = y. Then...

    2. The cardinality of a set is its "size". The cardinality of a finite set is the number of elements in it. Since A and B are finite sets having the same cardinality, |A| = |B| = n for some natural number n.
     
  4. May 3, 2008 #3
    How do you use |A| = |B| = n to prove f is onto?
     
  5. May 3, 2008 #4
    What have you tried so far? What happens if it isn't onto? In other words what happens if there is an element of B that is not mapped to by any element of A?
     
  6. May 4, 2008 #5
    Precise answer depends on used finite set definition.
     
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