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The exercise is taken from Rudin, principles of mathematical analysis, chapter 2 ex. 1.

Let $A$ a set and let also $B$ such that $A \cap B = \emptyset$ This implies:

$$ \emptyset = A \cap B \subseteq A, $$

For given $A$ the set $B$ can always be found, for example take $B = A^C$, the complement of $A$. Is such proof correct?

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  • $\begingroup$ Btw, it's usually good to repeat the question in the body of the text. $\endgroup$ – MCT Jul 4 '16 at 14:32
  • $\begingroup$ I think that's good! The proof by contradiction is worth looking up if you have the time, it's somewhat famous and also super short. $\endgroup$ – Lee Fisher Jul 4 '16 at 14:33
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Looks good. We can also do it like this:

$A \subset B$ is equivalent to saying $x \in A \implies x \in B$. This statement is vacuously true when $A = \varnothing$ (since there is no $x \in A$) and $B$ is any set. (even the empty set!)

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  • $\begingroup$ I was reading in the munkres book, topology, what vacousuly true means. But really it doesn't explain it in a nutshell what it is... $\endgroup$ – user8469759 Jul 4 '16 at 14:41
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    $\begingroup$ Vacuously true statements are usually of the type "for all $x$ satisfying some property, they also satisfy this property;" the statement is vacuous (but true! hence, vacuously true) since there are no $x$ satisfying the first property. In this case, there is no $x \in \varnothing$, so the statement is vacuously true. $\endgroup$ – MCT Jul 4 '16 at 20:55
  • $\begingroup$ It is still a bit twisted to understand. What I want to prove is the statement $x \in \emptyset \Rightarrow x \in A$, don't I want to prove that such statement is true? I take an $x$ such that $x \in A$, so the part $x \in A$ is true, while $x \in \emptyset$ is always false, this implies I have a statement of the form $F \Rightarrow T$ which literally by definition of implication is true. Is this the rationale beyond the proof? $\endgroup$ – user8469759 Jul 5 '16 at 6:48
  • $\begingroup$ "I take an $x$ such that $x \in A$, so the part $x \in A$ is true" That's where you're wrong. The statement $p \implies q$ is only false when $p$ is true but $q$ is false. The truth or falsity of $q$ is immaterial when $p$ is false, which it always is in this case. $\endgroup$ – MCT Jul 5 '16 at 13:53
  • $\begingroup$ Aaaaaaaaaaaaaah, I guess I've got it now. In general one would like to prove something like $p \Rightarrow q$ assuming $p$ is true so I necessarily need to prove that $p$ is true. In this case instead since $p$ is always false $q$ could be both true or false so I don't need to prove anything, am I right? $\endgroup$ – user8469759 Jul 5 '16 at 14:49
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You're missing the essential point. In fact, I'd say this is essentially circular. How do you know that the intersection of two sets is necessarily a subset of both of them? If $A \cap B$ is non-empty, then let $x \in A \cap B$ and, by definition, $x \in A \land x \in B$ which proves the theorem. However, if $A$ and $B$ are disjoint, then this proof isn't quite as obvious, is it?

To prove the exercise, note that $x \in \emptyset \implies x \in S$ means the same thing as $x \notin \emptyset \ \lor x \in S $. However, $\forall x, x \notin \emptyset$ and so the theorem is proved.

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  • $\begingroup$ Because, I suppose, you could define the empty set as the result of an intersection of two sets, $A$ and $B$, when they have no elements in common. $\endgroup$ – user8469759 Jul 4 '16 at 14:47
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$$\emptyset \subset A \equiv (\forall x) (x \in \emptyset \Rightarrow x \in A) \equiv \neg (\exists x) (x \in \emptyset \land x \notin A)$$

As $\emptyset$ is the empty set, $x \in \emptyset$ can never be true, i.e., $x \in \emptyset \land x \notin A$ is always false. Thus, $(\exists x) (x \in \emptyset \land x \notin A)$ is false and its negation is true. Hence, $\emptyset \subset A$ is true regardless of $A$.

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I don't like it.

It works in the context of some set $X$ serving as universe and containing $A$ as subset.

In my view preferable is: $\varnothing=A-A\subseteq A$.

Even more preferable: you can just prove that every element of $\varnothing$ is an element of $A$. This is vacuously true: no element of $\varnothing$ can be found that is not an element of $A$. Even stronger: no element of $\varnothing$ can be found at all.

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