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Let N be the set on Natural numbers. Let S={2,6,14} and T={6,8,11}.

Question: The intersection of S and T is nonempty

The correct answer is this:

$∃ x : x ∈ S. ∃ y : y ∈ T. x=y$

My answer is this:

$∃ x : x ∈ S ∧ x ∈ T$

Is my answer a sufficient way of saying the value x is in the intersection of S and T?

If not, why does the correct answer refer to two variables, x and y? Can x only represent the values in one set, even when they intersect?

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    $\begingroup$ Your answer is correct. $\endgroup$ – MJD Nov 17 '16 at 14:04
  • $\begingroup$ Note that the "correct" answer only quantifies over defined sets, whereas your answer quantifies over the entire universe. Depending on your text, unrestricted quantification may not be allowed. $\endgroup$ – DanielV Nov 17 '16 at 23:42
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Your answer is correct ... if you add some parentheses:

$\exists x (x \in S \land x \in T)$

This is logically equivalent to the provided answer, and so no, you don't need a different variable just because you have different sets; it is the same object that is in the intersection, so you can indeed do exactly what you do.

Finally, your statement is not quite saying 'x is the intersection of S and T' (because for different S and T they may have more than 1 element in common), but it is saying that 'x is in the intersection of S and T', and hence that the intersection is not empty.

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  • $\begingroup$ is there any reason to refer to 2 variables and 2 quantifiers instead of one? $\endgroup$ – Francis Nov 17 '16 at 14:02
  • $\begingroup$ Your version and the OP's version are just the same thing written in different syntax. $\endgroup$ – Hurkyl Nov 17 '16 at 14:03
  • $\begingroup$ The provided answer doesn't use parentheses either. Notations vary, and it is correct for Francis to use the same notation that the instructor is using, and not some different notation. $\endgroup$ – MJD Nov 17 '16 at 14:04
  • $\begingroup$ I thnk the provided answer has no parenthesis since it requires no parentheses. Think of it as $\exists x \in S \: \exists y \in T x = y$. Some syntactical notations are fine with that. But I have not seen any notation that is fine with $\exists x x \in S \land x \in T$ $\endgroup$ – Bram28 Nov 17 '16 at 14:09
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    $\begingroup$ @Francis No, you don't need two variables just because you are dealing with two sets. Just added that to my answer. $\endgroup$ – Bram28 Nov 17 '16 at 14:13

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