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Another Set Theory Proof(s)

  1. Nov 18, 2009 #1

    (i)A[tex]\subseteq[/tex]B [tex]\Leftrightarrow[/tex] A[tex]\cup[/tex]B = B
    (ii) A[tex]\subseteq[/tex]B [tex]\Leftrightarrow[/tex] A[tex]\cap[/tex]B = A


    For subsets of a universal set U prove that B[tex]\subseteq[/tex]A[tex]^{c}[/tex] [tex]\Leftrightarrow[/tex] A[tex]\cap[/tex]B = empty set. By taking complements deduce that A[tex]^{c}[/tex][tex]\subseteq[/tex]B [tex]\Leftrightarrow[/tex] A[tex]\cup[/tex]B = U. Deduce that B = A[tex]^{c}[/tex] [tex]\Leftrightarrow[/tex] A[tex]\cap[/tex]B = empty set and A[tex]\cup[/tex]B = U.

    Can't wrap my head around the last question at all. The i and ii seem simple but I'm just not getting it to work.
  2. jcsd
  3. Nov 18, 2009 #2
    (i) If A is empty, claim is true trivially. If it's not, then take an element of A, [tex] x \in A [/tex].

    Suppose A is a subset of B. What does this mean for x? Use the definition of the cup operation :) Then suppose [tex] A \cup B = B [/tex] and do the same.

    For (ii) you might want to assume that A is not empty because the thing you're trying to prove does not generally hold if both A and B are empty (mathematicians are weird...)

    For the last bit, use again some element of b, [tex] y \in B [/tex]. Show that if [tex] y \in A^c [/tex] then it can't be in A.
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