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Functions, intersections, and unions

  1. Jan 18, 2005 #1
    Okay I need to show that for a function g,

    g(A n B) is a subset of g(A) n g(B) where A and B are sets and n means intersection.

    I also need to show something similar about the union but I'm not given the relation. I have to figure that out myself. Any pointers on how to get started on either of these?
     
  2. jcsd
  3. Jan 18, 2005 #2

    AKG

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    To show that:

    [tex]g(A \cap B) \subseteq g(A) \cap g(B)[/tex]

    it suffices to show that:

    [tex]x \in g(A \cap B) \Rightarrow x \in g(A) \cap g(B)[/tex]

    This basically says that if you have any sets X and Y, and X is inside Y (i.e. X is a subset of Y), then any element inside X must obviously be inside Y. So, start by assuming [itex]x \in g(A \cap B)[/itex] and deduce [itex]g(A) \cap g(B)[/itex]. It's pretty clear. If [itex]x \in g(A \cap B)[/itex], then [itex]x \in g(A)[/itex], since [itex](A \cap B) \subseteq A[/itex]. Similarly, if [itex]x \in g(A \cap B)[/itex], then [itex]x \in g(A)[/itex]. So we have that if x is in [itex]g(A \cap B)[/itex], then x is in both g(A) and g(B). Obviously, this means that x is in [itex]g(A) \cap g(B)[/itex], and the proof is done.

    I'll let you prove the union one, as you should know how to do it now. Although I'm not sure, my guess would be that g(A) U g(B) is a subset of g(A U B). Note how it's kind of like the "backwards" version of the previous proposition. Intersection and Union have that type of relationship, i.e. if you have a proposition with unions in it, then you can sort of "invert" things and change unions to intersections to get another true proposition.

    For example, if we let A' denote the complement of A, then:

    [tex]A\prime \cap B\prime = (A \cup B)\prime[/tex]

    The intersection of complements is the complement of a union. If you take the complements first, then you take the intersection of the sets. If you "invert" the order, and take the complement last, then you union the sets first.
     
  4. Jan 18, 2005 #3
    Oh and one more thing,

    I know that (A n B) n C = A n (B n C)
    and that A n (B u C) = (A n B) U (A n C).

    I remember those properties from another class I took but I cannot remember what they are called. Can someone tell me this?
     
  5. Jan 18, 2005 #4
    first one is associative, second is distributive (just like algebra),,, there are many more including DeMorgan's laws
     
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