1. Sep 2, 2010

### ibc

Hello

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?]

Last edited by a moderator: Apr 25, 2017
2. Sep 2, 2010

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.

3. Sep 2, 2010

### ibc

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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