Recent content by AKG

@Aziza, what do you mean by "any t \in \mathbb{R}". If your f is a function, then for a given input x, you need to specify a unique output f(x). E.g. f(1) can only be one point in the plane, you can't have f(1) = (1, 0), and also f(1) = (1,1), and also f(1) = (1, -\pi), and so on. Since I...

If B satisfies s, then it doesn't satisfy -s, hence A doesn't satisfy -s, hence A satisfies s.

It definitely uses extra axioms, indeed it can't follow from ZFC alone since it would contradict CH, which is consistent with ZFC. Whenever you see something like: Theorem: <Statement> It means the <Statement> is a theorem of ZFC. Whenever you see: Theorem: (<Axiom(s)>) <Statement> It...

One can give an explicit set of instructions to construct the model, but the process is rather indirect and might not be what you're looking for. That is, there's a positive answer to your question about whether there is a prescription, you just might not like the taste of it. At any rate...

EDIT: Oops, sorry, when I said it was clear that the X_i+X_j are clearly iid I was mistaken. I'm not sure that they're not independent, but if they are independent, it certainly isn't clear. -------- The CDF of this random variable, let's call it X, is given by: a \mapsto P(X < a)...

Assume choice and let (P,<) be a poset satisfying the hypotheses of Zorn's Lemma, but failing to have a maximal element. Let X be the collection of subsets of P which are well-ordered by <. Let F : X \to \mathcal{P}(P)\setminus \{\emptyset\} be such that F(x) is the set of all strict upper...

ZFC is a first order theory, where have you been seeing second order formulations? I'm sure every one of the top Google results for "ZFC axioms" will give you a first order formulation. In particular, Wikipedia.

Yes, they're all at least countable, because they all have an empty set, a set containing just the empty set, a set containing just the set containing just the empty set, etc. Your argument using Replacement doesn't work, since there's no guarantee that two different instances of Replacement...

Correct. Another simple argument is to do the following construction: X_0 = Hull(\{\beta\},L_{\alpha}) \beta_0 = \min\{\gamma : X_0 \subset L_\gamma\} X_{n+1} = Hull(L_{\beta_n},L_\alpha) \beta_{n+1} = \min\{\gamma : X_n \subset L_\gamma\} L_{\beta'} = \bigcup X_n Here Hull(X,M) denotes some...

"The union of elementary submodels is itself an elementary submodels" is only true when those submodels are elementary substructures of one another. This means more than just the fact that L_{\alpha_n} \subset L_{\alpha_{n+2}} and L_{\alpha_n} \equiv L_{\alpha_{n+2}}, it requires that the...

I recommend just computing the determinant and doing some simple algebra to factor the resulting expression.

This is not possible for n>2. If x_i \sim \mathrm{Unif}(0,1) for i = 1, \dots, n, then 1-x_n \sim \mathrm{Unif}(0,1) as well. But if they sum to 1, then this means x_1+\dots +x_{n-1} \sim \mathrm{Unif}(0,1). But \mathbb{E}[x_1+\dots x_{n-1}] = \frac{n-1}{2} \neq \frac 12

Hmmm, what you're saying definitely isn't right. You're not mentioning \phi at all. We've got our set of natural numbers \mathbb{N} = \{ 0, 1, 2, \dots \}. And then we've got some other random set a. Now we can talk about functions from \mathbb{N} to a. We can imagine the collection of all...

They construct the fixed point of \phi. They call this fixed point L_0, and they give you a recursive definition of this function, although admittedly their description is a bit confusing because it involves this intermediate definition of the functions \phi_n. Let's make sure you understand...

What does the fact that x\in U with U open tell you? Based on what that gets you, you should be able to think of a candidate for the desired open neighbourhood containing [x] and contained in \phi(U).