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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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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. –  Shahab Oct 9 '12 at 1:57
    
@Shahab I don't understand... for a discrete math course should I be expecting that? –  Rice Newman Oct 9 '12 at 2:00
    
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. –  Shahab Oct 9 '12 at 3:32

2 Answers 2

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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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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2  
"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. –  Austin Mohr Oct 9 '12 at 1:52
    
@AustinMohr - indeed, this is exactly what I meant. thank you for your comment –  Belgi Oct 9 '12 at 1:53
    
How come on this website, it says otherwise (6th diagram) purplemath.com/modules/venndiag2.htm –  Rice Newman Oct 9 '12 at 2:26
    
@RiceNewman - you want the second one. why the 6th ? –  Belgi Oct 9 '12 at 2:28
    
Yes, the second or the 6th. They are both showing ^ –  Rice Newman Oct 9 '12 at 2:30

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