Undergraduate mathematical logic questions

AI Thread Summary
The discussion revolves around difficulties in understanding concepts from Y. Manin's "Introduction to Mathematical Logic." One participant seeks clarification on proving Mostowski's proposition regarding the union of subsets in the Von Neumann universe, questioning why a set being a subset of a union implies it must belong to one of the individual sets. Another participant addresses the axioms of Boolean algebras, specifically the claim that the value of simple tautologies must be 1, expressing confusion over proving the identity a' ∨ a = 1. The conversation highlights the need for clearer axiomatic foundations and conditions for these logical statements. Overall, the thread emphasizes the complexities of mathematical logic and the importance of precise definitions and proofs.
ibc
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Hello
I'm reading Y. Manin's http://books.google.co.il/books?id=...resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false" and I've been having some difficulties. So I'm asking for help.

1. On page 64 of the book, (prooving Mostowski proposition). We have a subset N of the Von neumann universe, and N_\alpha \subset N for all cardinals \alpha. Now we want to prove that \cup N_\alpha= N, his first step is, assume otherwise and assume there is X\in N\setminus \cup(N_\alpha) such that X \cap (N\setminus \cup N_\alpha) = \emptyset then X\subset \cup N_\alpha thus there is some \alpha_0 s.t X\subset N_\alpha_0
Now, that last part I don't understand. why the fact that a set is a subset of a union of sets, it must be a subset of one of them? (which is clearly not true as a general argument, but why is it true here?)

2. On page 52, he presents Boolean algebras with axioms
(A^')^' = A
\vee \wedge are associative commutative and distributive
(a \vee b )^' = a \wedge b (a \wedge b)^' = a \vee b
a \wedge a = a \vee a = a
1 \wedge a = a 0 \wedge a = a

later he claims that for any map from a set of formulas to the boolean algebra, the value on the simple tautologies must be 1
but I don't see why it is so. isn't there something missing? for example, I can't see how to prove that a^' \vee a = 1.

Thanks

[Sorry for all these Latex oddities, does anyone know how to exit the "uppercase" mode?]
 
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ibc said:
Now, that last part I don't understand. why the fact that a set is a subset of a union of sets, it must be a subset of one of them? (which is clearly not true as a general argument, but why is it true here?)
It's because you're considering the union of a chain of sets (that is; for any two of the sets, one is contained in the other).

ibc said:
2. On page 52, he presents Boolean algebras with axioms
(a')' = a
\vee \wedge are associative and commutative, and distributive over each other
(a \vee b )' = a' \wedge b'; (a \wedge b)' = a' \vee b'
a \wedge a = a \vee a = a
1 \wedge a = a; 0 \vee a = a

later he claims that for any map from a set of formulas to the boolean algebra, the value on the simple tautologies must be 1
but I don't see why it is so. isn't there something missing? for example, I can't see how to prove that a' \vee a = 1.

Thanks

[Sorry for all these Latex oddities, does anyone know how to exit the "uppercase" mode?]

Just use ' instead of ^'. Also your axioms were incorrect; I've fixed it above. To prove a' ∨ a = 1, you must show that it suffices to show that (a' ∨ a) ∧ b = b for all b. I'd have to think about it more to prove it from those axioms, since I'm used to a different set of axioms.
 
adriank said:
It's because you're considering the union of a chain of sets (that is; for any two of the sets, one is contained in the other).



Just use ' instead of ^'. Also your axioms were incorrect; I've fixed it above. To prove a' ∨ a = 1, you must show that it suffices to show that (a' ∨ a) ∧ b = b for all b. I'd have to think about it more to prove it from those axioms, since I'm used to a different set of axioms.

I don't understand. Take N= natural numbers
N\subset \cup N_n
where N_n is all the natural numbers up to n. it is a chain, and obviously N is not contained in any finite union. (and obviously there are similar less "dramatic" examples (where the subset is not all the union))

yes, obviously it should be as you wrote it. and yes, we probably should assume that it's sufficient to prove that (although the book doesn't mention any iff condition about this property of "1", but without it it seems even less likely that such axioms are enough to conclude the wishful conclusion)
I failed to prove that (a' ∨ a) ∧ b = b for all b using these axioms. I just don't see how it's possible, it seems like the only property of "1" the axioms give us is the one we are trying to prove, so we've got nothing to work with.
 
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