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Here’s a problem from a prior test. Didn’t do too hot on it so I redid it. How did I do?

If A and B are any sets, show that A∩B and A\B are disjoint and that A=(A∩B)∪(A\B).

My work:

Show that A∩B and A\B are disjoint: Assume x∈(A∩B)∩(A\B) then x∈(A∩B) and x∈(A\B) ⇒x∈A,B and x∈A and x∉B ⇔ So ∅=(A∩B)∩(A\B)

Show that A=(A∩B)∪(A\B) Assume x∈(A∩B)∪(A\B) then x∈(A∩B) or x∈(A\B) ⇒(x∈A,B) or (x∈A and x∉B) Either case x∈A

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    $\begingroup$ You have shown that $(A\cap B)\cup(A\setminus B)\subseteq A$. You still need to the reverse inclusion. $\endgroup$
    – Tim Raczkowski
    Sep 28, 2015 at 22:45
  • $\begingroup$ If x∈A and x∉B then x∉(A∩B) and x∈(A\B). So x∈(A∩B)∪(A\B). Then do x is an element of B... $\endgroup$
    – Crossing the Abyss
    Sep 28, 2015 at 22:56
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    $\begingroup$ Yes. That's it. The proof is complete. $\endgroup$
    – Tim Raczkowski
    Sep 28, 2015 at 22:57
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    $\begingroup$ In the first of the two lines that start with "Show ..." I think you need $x\in(A\cap B)\color{blue}{\cap}(A\setminus B)$ not x∈(A∩B)∪(A\B), if you want to show the two are disjoint $\endgroup$
    – Marconius
    Sep 29, 2015 at 1:36

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