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I need help with this proof:

Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.

My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.

Where do I go from here?

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    $\begingroup$ Please, try to make the title of your questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. $\endgroup$ – Lord_Farin Oct 25 '13 at 22:22
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Hint:

$$x\in A\cap (B\setminus A)\implies x\in A\;\text{ and also }\;x\in B\setminus A\implies$$

$$x\in A\;\text{and also}\;\left(x\in B\;\text {but}\;x\notin A\right)\implies\ldots$$

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    $\begingroup$ Is it like a contradiction? $x\in A$ but $x\notin A$. $\endgroup$ – user87274 Oct 25 '13 at 21:50
  • $\begingroup$ So our assumption that $x\in A\cap (B\setminus A)$ is wrong and so $A\cap (B\setminus A)$ must be empty. $\endgroup$ – user87274 Oct 25 '13 at 22:00
  • $\begingroup$ Exactly @AJR: if there's an element in $\;A\;$ then that element isn't in $\;A\;$ , contradiction! Thus, no elements are possible in that intersection and it is then empty. $\endgroup$ – DonAntonio Oct 26 '13 at 8:14
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Hint for a more "algebraic" solution:

$$ A \cap (B \setminus A) = A \cap (B \cap A^c) = A \cap (A^c \cap B) = (A \cap A^c) \cap B = \ldots.$$

By the way, the problem seems a bit strange for asking you to show that a given set is a subset of the empty set. In general, to show $X \subseteq Y$ you would let $x \in X$ and prove that $x \in Y$. So to show that $X \subseteq \emptyset$ you would let $x \in X$ and then prove that $x \in \emptyset$. Of course "$x \in \emptyset$" is absurd, so this method amounts to letting $x \in X$ and proving absurdity (a contradiction.) Note that a contradiction implies everything, including the desired conclusion that $x \in \emptyset$.

An equivalent but more natural formulation of the question would simply ask you to show that the given set was empty. After all, the only subset of the empty set is the empty set itself.

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  • $\begingroup$ Yeah the question is a bit weird. $\endgroup$ – user87274 Oct 25 '13 at 22:25
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Here is a slightly more calculational rendering of the proof in DonAntonio's answer, but using only equivalences: translate from the set level to the logic level, expanding the definitions and simplifying along the way.

In this case, for every $\;x\;$, \begin{align} & x \in A \cap (B \setminus A) \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$"} \\ & x \in A \land x \in B \setminus A \\ \equiv & \;\;\;\;\;\text{"definition of $\;\setminus\;$"} \\ & x \in A \land x \in B \land x \not\in A \\ \equiv & \;\;\;\;\;\text{"simplify using contradication"} \\ & \text{false} \\ \end{align}

By the definition of $\;\varnothing\;$, this proves $\;A \cap (B \setminus A) = \varnothing\;$, and therefore also $\;A \cap (B \setminus A) \subseteq \varnothing\;$.

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