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Show $(A \cup B)\setminus(A \cap B) = (A\setminus B) \cup (B\setminus A)$.

What I have so far...

This is (A or B) and (A and B)' = (A and B') or (B and A')

(A or B) and (A' or B') = (A and B') or (B and A')

((A or B) and A') or ((A or B) and B') = (A and B') or (B and A')

I feel like I'm just getting farther away. There must be something simple I'm missing.

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  • $\begingroup$ Try using venn diagrams to visualise this can help with the proof. $\endgroup$ – Kaladin Oct 13 '15 at 22:37
  • $\begingroup$ I am. It helps, but I feel that it makes me want to jump to make claims which I should be able to make by simply knowing general set algebra. $\endgroup$ – user2684794 Oct 13 '15 at 22:40
  • $\begingroup$ Alright. I figured it out. $\endgroup$ – user2684794 Oct 13 '15 at 22:48
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As Simon's answer points out, often you want to construct a so-called "element-chasing" proof to show that $S=T$ by showing $S\subseteq T$ and $T\subseteq S$.

However, I would only use an element-chasing proof if you can't "squeeze out" the equality by using some basic set algebra. Assuming you know about what the complement of a set is, distributivity, etc., see if you can follow the proof below (more steps than usual have been added to emphasize clarity, but let me know if a step or two does not make sense): \begin{align} \text{LHS}&=(A\cup B)\setminus(A\cap B)\\ &= (A\cup B)\cap(A\cap B)^C\tag{definition}\\[0.5em] &= (A\cup B)\cap(A^C\cup B^C)\tag{De Morgan}\\[0.5em] &= [(A\cup B)\cap A^C]\cup[(A\cup B)\cap B^C]\tag{distrib.}\\[0.5em] &= [(A\cap A^C)\cup(B\cap A^C)]\cup[(A\cap B^C)\cup(B\cap B^C)]\tag{distrib.}\\[0.5em] &= [\varnothing\cup(B\cap A^C)]\cup[(A\cap B^C)\cup\varnothing]\tag{$S\cap S^C=\varnothing$}\\[0.5em] &= (B\cap A^C)\cup(A\cap B^C)\tag{simplify}\\[0.5em] &= (A\cap B^C)\cup(B\cap A^C)\tag{commutativity}\\[0.5em] &= (A\setminus B)\cup(B\setminus A)\tag{definition}\\[0.5em] &= \text{RHS}. \end{align} Are all of those steps clear? Can you see how the proof above, despite seeming somewhat long winded, is exactly a good bit simpler than cranking out two proofs (one for each direction)?

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  • $\begingroup$ Yup. That's the route I took. $\endgroup$ – user2684794 Oct 13 '15 at 23:32
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Hint:

In general to show that two sets, say $S$ and $T$, are equal you want to show that $S \subset T$ and $T \subset S$. In other words $x \in S \Rightarrow x \in T$ and $x \in T \Rightarrow x \in S$.

Hence what you want to show is that

$$1. \quad x \in (A \cup B)\backslash (A \cap B) \ \Rightarrow x \in (A\backslash B)\cup(B\backslash A)$$ and

$$2. \quad x \in (A\backslash B)\cup(B\backslash A) \ \Rightarrow x \in (A \cup B)\backslash (A \cap B) $$

Try writing arguments for these two statements. A Venn diagram will also help you see what is going on here.


Here's the beginning of a possible argument for statement 1:

If $x \in (A \cup B)\backslash (A \cap B)$ then $x \in A$ or $x \in B$. However if $x \in A$ then $x \not\in B$ as $x \not\in A \cap B$. Similarly if $x \in B$ then ...

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