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I have been exploring more about set theory beyond my textbook and I have ran into something I couldn't explain. Can you use logical conjunction/disjunction on sets and are they the same as union/intersection?

A $\bigcap$ B

A $\wedge$ B

where A and B are sets. Are these equivalent? What does the disjunction or two sets mean?

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  • $\begingroup$ Note that if $S$ is some set, then $\mathcal{P}(S)$, the union, intersection and complementation forms a Boolean algebra. Similarly, the set of propositions with $\lor$, $\land$ and $!$ forms a Boolean algebra which may be established to be isomorphic. $\endgroup$ – Shahab Oct 9 '12 at 1:57
  • $\begingroup$ @Shahab I don't understand... for a discrete math course should I be expecting that? $\endgroup$ – Rice Newman Oct 9 '12 at 2:00
  • $\begingroup$ I am not sure whether you have studied Boolean algebra, but if you haven't just note that there is a transformation which transforms conjunction/disjunction between propositions to union/intersection between sets nicely. This is really besides the point of your question and hence I included it just as a comment. $\endgroup$ – Shahab Oct 9 '12 at 3:32
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The relation is that $$x\in A\cap B\iff (x\in A)\wedge (x\in B)$$

You can't write something like $$A\wedge B$$ it is not defined.

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    $\begingroup$ "You can't write something like..." As long as you provide a definition, you can write whatever you want. I think you mean to say $A \wedge B$ has no standard meaning. $\endgroup$ – Austin Mohr Oct 9 '12 at 1:52
  • $\begingroup$ @AustinMohr - indeed, this is exactly what I meant. thank you for your comment $\endgroup$ – Belgi Oct 9 '12 at 1:53
  • $\begingroup$ How come on this website, it says otherwise (6th diagram) purplemath.com/modules/venndiag2.htm $\endgroup$ – Rice Newman Oct 9 '12 at 2:26
  • $\begingroup$ @RiceNewman - you want the second one. why the 6th ? $\endgroup$ – Belgi Oct 9 '12 at 2:28
  • $\begingroup$ Yes, the second or the 6th. They are both showing ^ $\endgroup$ – Rice Newman Oct 9 '12 at 2:30
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They are closely related. One of the main motivations behind the notion of a set (or more generally of a class) is as an object that corresponds to a logical predicate: the set/class is an aggregation whose elements are precisely the objects satisfying the predicate.

If $P$ is a unary predicate, and I use the notation $[P]$ to denote the class corresponding to $P$: i.e. the class satisfying

$$x \in [P] \Leftrightarrow P(x)$$

then we have

$$[P \wedge Q] = [P] \cap [Q]$$

so we see the close relationship between $\wedge$ and $\cap$.

But for sets $S$ and $T$, $S \wedge T$ doesn't really make sense. Notation like this is would appear, however, if you were working in a lattice whose elements are sets: in this case, $\wedge$ is not meant to be viewed as an operation on sets, but as an operation on lattice elements (which just happen to be sets).

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"Can [one] use logical conjunction/disjunction on sets?"

No, one cannot. Conjunction and disjunction are operations over two propositions/sentences in models of Propositional/Sentential Logic. Boolean Algebra is one of such models. These logical operations are not applicable to sets of elements.

What you can do is relate a set such as "all teachers" (T) and "all Americans" (A) with propositions such as "John is a teacher" (T(J)) and "John is American" (A(J)). As pointed by others above, this results in expressions such as "someone is one of the teachers if and only if this person is a teacher" (x ∈ T ⟺ T(x)). If you combine them, you get something like "someone is one of the Americans who are teachers if and only if this person is American and this person is a teacher" (x ∈ A ∩ T ⟺ A(x) ∧ T(x)).

NOTE: These operations are not necessarily what happens in our daily interactions. For instance, what is meant by "American teachers" is usually "all people who teach in the US" independent of their nationality. This mismatch only means that it is not trivial to map human adult language to logical expressions, not that some logics are wrong or unusable in language processing.

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