Actually, here's an interesting relation between set-inclusion and material implication. Both satisfy transitivity and reflexivity. If two sets include each other, then they are equal, and if two sentences materially imply each other, then they are materially equivalent.
honestrosewater said:
Doh! How did I miss this? You mean in assigning truth-values? If so, aren't they still different since they assign values that aren't a part of the language or structure or theory or however you want to look at it?
Any operation * on a set U can be seen as a function:
* : U x U --> U
If S and T are elements of U, then we normally write operations with infix notation so *(S, T) is denoted as S*T, and S*T is an element of U. So if * is the union operation, and U is the set of all sets, then S and T are sets, and S*T is the union of S and T, which is also a set hence also a member of U. If U is the set of propositions and * is conjugation, then S*T is just another proposition. Relations, if you want to treat them as function from U x U to something, it would be to the set of truth values {t, f}.
S \subset T
is not a set, it's either yes (S is contained in T) or no. A relation then is a map:
R : U x U --> {t, f}
However, it is "standard" to think of relations as sets of ordered pairs. If R is a relation on U, and s and t are elements of U, then we say (s, t) is in R iff sRt. If R is the set-inclusion relation, then
R = {(s, t) : s, t in U; x in s implies x in t}
An operation is a function, and a function can also be thought of as a set of ordered pairs, but it would be completely different kinds of ordered pairs. The ordered pairs would consist of arguments in the first component and the corresponding value of the function in the second component. So a function f : U x U --> U would be of the form:
f = {((s, t), f(s, t)) : s, t in U}
Note that every element of f is an ordered pair,
and the first component of each such pair is itself another ordered pair. If U is the set of propositions and f is the implication operation, then
f = {((p,q), (p --> q)) : all propositions p, q}
If you want to look at it as what assigns what, then relations always assign truth values to a pair of objects, and binary operations always assign an object (of the same type) to a pair of objects. Logical-connectives, set union, set intersection, etc. are all binary operations. Set inclusion is a binary relation.
I can't see how set theory and propositional logic have much in common. Maybe I'm just thinking about this the wrong way, but they don't seem to have the same kind of structure. In propositional logic, the individuals are assigned truth-values. What would it mean - and how would it work - to assign a truth-value to a set? I did notice in another thread that connectives do behave somewhat like relations in seeming to have something like reflexive, symmetric, and transitive properties. But connectives just don't seem to work as relations; I can't quite put my finger on the reason. Hm, it might be fun to get into this some more if anyone else is interested.
This would be interesting. A is a set, and on it's own, it's just an expression, not an equation like A = B which could be assigned a natural truth value. Also, relations and operations are different, but as I suggested at the start of my post, set-inclusion does have some similarities to implication.
However, I've thought about it a little, it seems tough. We need to find an analogy to truth, and an analogy to inference. Every sentence in propositional logic can be written using atomic sentences, conjunction, and negation, so if we correlate sets with propositions, then we can make any proposition-set with "atomic" sets, intersection, and complement. But I still can't tell how to associate a truth value with a set (although the universal set should be tautology or logical truth and the empty set should be contradiction or logical falsity) and I can't figure out what type of set-relation would stand for inference. I'll have to think about it some more.
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