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An exercise asks me to prove the following: $$A \cap (B-A) = \varnothing$$ This is what I did:

I need to prove that $A \cap (B-A) \subseteq \varnothing$ and $\varnothing \subseteq A \cap (B-A)$

The second one seems to be obvious by definition (not sure if it is fine to say that in a test).

But the first one goes like this: $$x \in A \cap (B-A)$$ $$x \in A \land x \in (B-A)$$ $$x \in A \land (x \in B \land x \notin A)$$ $$(x \in A \land x \notin A) \land x \in B$$ $$(x \in A \land x \notin A)$$ Since there is a contradiction... err... the proof... is good.

Okay, that's my problem. I think that the procedure is fine, but clearly I am unable to word it out (well, I don't even know if it is valid at all). Is it valid? How can I... "prove that my proof is correct"? I don't really see this contradiction as an obvious indicator of $A \cap (B-A) \subseteq \varnothing$

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    $\begingroup$ Just a small comment: your proof actually shows $A\cap (B-A)=\emptyset$ (no need to do the "other" set containment) $\endgroup$ – icurays1 Dec 4 '12 at 5:47
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Your logic is correct, but at your level of detail I would write the argument more or less as follows.

Suppose that $x\in B\setminus A$. Then by definition $x\notin A$, so $x\notin A\cap(B\setminus A)$. And if $x\notin B\setminus A$, then certainly $x\notin A\cap(B\setminus A)$, so for every $x$ we know that $x\notin A\cap(B\setminus A)$, i.e., $$A\cap(B\setminus A)=\varnothing\;.$$

Note that words usually make better explanatory ‘connective tissue’ for an argument than a bunch of arrows.

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Yes this is perfectly valid.

You derive a contradiction from the assumption that there is an $x$ satisfying $$x \in A \cap (B\setminus A),$$ so there must be no $x$ satisfying this inclusion and the set is empty.

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Everything is correct. The contradiction is that the element $x$ you found is both in $A$ and not in $A$, which is impossible. Thus, there is no $x\in A\cap(B-A)$, which implies it is empty.

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Directly:

$$A \cap (B-A) = A \cap (B \cap \overline A) \\ = (A \cap \overline A) \cap B \\ = \varnothing \cap B \\ = \varnothing $$

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