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Update : Should have left the Arithmetic out of this question, the new modified question is posted here : $\wedge,\cap$ and $\vee,\cup$ between Logic and Set Theory always interchangeable?

Although have not seen this anywhere stated, but so often mentally swapping the frameworks of Logic, Set Theory and Arithmetic following operation $\wedge,\cap,\times$ and $\vee,\cup,+$ are seem to be always interchangeable (Isomorphic?).

My question is are $\wedge,\cap,\times$ the same thing and just different symbols are being used depending on the framework? same question regarding $\vee,\cup,+$

PS: In arithmetic setting we get the case of $[0, \text{any number other than 0}] \equiv [false , true] $

For example De Morgan's laws for Sets and Logic just becomes the distributive and associate laws.

What about infinite cases? does this type of intuition break down between Arithmetic, Logic and Set theory?

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In Set Theory, $\cap$ and $\cup$ distribute over each other: $a\cap(b\cup c) = (a\cap b)\cup (a\cap c)$ and $a\cup(b\cap c) = (a\cup b)\cap (a\cup c)$. In arithmetic, $\times$ distributes over $+$, but $+$ does not distribute over $\times$; that is, $a\times(b+c) = (a\times b)+(a\times c)$, but in general, we do not have $a+(b\times c) = (a+ b)\times (a+ c)$. So even in this very basic setting, you have that $\cap$ and $\times$ have different properties relative to $\cup$ and $+$, respectively. –  Arturo Magidin Feb 18 '12 at 23:11
@ArturoMagidin : That is the counter example I needed, can you please post it as an answer so I can accept it? Thank you –  Arjang Feb 18 '12 at 23:23
But still $x \in A \cap B$ is the same as $(x\in A) \wedge (x \in B)$ and the same for infinite cases (but I have never seen an infinite conjunction contrary to infinite intersection). Similarly for $\cup, \vee$: $x \in \{y: \phi(y)\} \cup \{y: \psi(y)\}$ is the same as $\phi(x) \vee \psi(x)$. –  savick01 Feb 18 '12 at 23:39
@savick01: Infinite conjunctions and disjunctions are sometimes seen, e.g., $\bigwedge\limits_{n\in\omega}(x\ne c_n)$. –  Brian M. Scott Feb 18 '12 at 23:51
I've seen "$+$" used to denote the symmetric difference between sets, since that corresponds to mod-2 addition of indicator functions. So that's certainly not the same as "$\cup$". –  Michael Hardy Feb 19 '12 at 0:05
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2 Answers

up vote 9 down vote accepted

As requested; (I think the question is a bit confused and fuzzy as it stands, but...)

In set theory, $\cap$ and $\cup$ distribute over each other: $$a\cap(b\cup c) = (a\cap b)\cup (a\cap c)\qquad\text{and}\qquad a\cup(b\cap c) = (a\cup b)\cap (a\cup c).$$

However, in arithmetic (by which I understand the Peano arithmetic or its extensions to $\mathbb{Z}$, $\mathbb{Q}$, etc), while $\times$ distributes over $+$, $+$ does not distribute over $\times$. We always have $a\times(b+c) = (a\times b)+(a\times c)$, but almost never have $a+(b\times c) = (a+b)\times(a+c)$.

So $\cap$ and $\cup$ in set theory have properties, relative to one another, that are different from the properties of $\times$ and $+$ in arithmetic.

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Thank you Arturo, including arithmatic in this question was wrong, the similarity is extreeme in Logic and Set Theory –  Arjang Feb 19 '12 at 0:29
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"+" isn't like "$\lor$". (x+x)=(2*x). On the other hand, in classical logic, (x $\lor$ x)=x. Specifically, (1+1)=2, while (1 $\lor$ 1)=1.

It comes as more useful to compare "∧,∩, min" and "∨,∪, max". Logical and set-theoretic operations behave more like the minimum of two numbers, and the maximum of two numbers than anything else (and max and min are dual operations if we have (1-x) as the complement of x).

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agreed, it was a mistake to include arithmatic the new question is posted here : math.stackexchange.com/questions/110760/… –  Arjang Feb 19 '12 at 0:27
Not "more than anything else". They're just all lattice operations, and there are plenty of other examples. Also, $+$ is a bit like $\vee.$ There are differences, but there are similarities too. –  user23211 Apr 3 '12 at 23:59
@user2311 A lattice L can get defined as a partially ordered set where for all x, for all y in L, sup{x, y} exists and inf{x, y} exists. Do you deny that max is closer to sup than anything else, and that min is closer to inf than anything else? –  Doug Spoonwood May 23 '13 at 20:36
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