Recent content by oliphant

There question then asks what the maximum rank for each of the 3 sets we defined would be (taking alpha, beta as ranks for a, b). for a) would it simply be alpha, as in when beta is the empty set, right? What sort of reasoning could I use for b?

Brilliant, thank you very much!

Oops, that's what I wrote down first of all (well a\b = \{x \in a | x \notin b\} for a) but then decided to over complicate it and confuse matters. So if we think of the function as a set of ordered pairs (s,t) (where the same a cannot give different bs). I actually have a definition for the...

Homework Statement We're asked to prove that a few constructions of the sets a,b are themselves sets, stating which axioms we use to do so. a) a\b b)the function f:a->b c)the image of f Homework Equations The following standard definitions of axioms of construction...

We were actually given a proof using the Axiom of Choice by applying it to the set {f^-1 (b) | b \in B} and defining g(b) = c(f^-1(b)) What if I was to define g(b) = \{a \in A | f(a) = b\}? Then g(b) would be a subset of A and so well ordered, and I think I can finish the proof from there.

Homework Statement Without using the Axiom of Choice, show that if A is a well-ordered set and f : A -> B is a surjection to any set B then there exists an injection B -> A. Homework Equations The Attempt at a Solution I was wondering if the existence of the surjection from a well...

I see, I think I just needed the argument about density before to get it straight in my head. I should be able to form a coherent proof from all this. Thank you very much for your help, it's greatly appreciated.

Well we know X has a first and last element by the earlier contradiction, and is not dense. So it's finite. Correct?

Ok, hammer is approaching the head of the nail. We haven't been explicitly told that well ordered sets cannot be dense. But, I think from the proposition "every element in X except the last one has an immediate successor" we can prove that X is not dense. Starting with an xi in X, we know...

So should I try and prove that X is not dense? OR, is there a way I could adapt the isomorphism method? I found a lemma that states if X is well ordered, each element of X except the first is a successor of some other element of X and X has no greatest element then (X, <) is isomorphic to (N...

Hm, well backtracking from the follow on question for a second, it seems that if we know the set has a minimum and a maximum element we can't necessarily say it is finite. Surely we have to consider the case where X is dense? We'd need to prove that there exists a bijection from X to...

Wow, that was much simpler than I thought it would be. Teach me to judge a question by how many marks it's worth. There's a follow on question "Deduce that any strictly descending chain of ordinals has finite length." Assume the chain is infinite α > β > .. But we know α > β > .. > 1...

After sleeping on it, I realize I was thinking of X as an interval, not a set. Hang on, if x0>x1>x2>... doesn't terminate it doesn't have a least element. But we know X is a subset of a well ordered set (A,≤), so is well ordered by ≤. So ... x2 ≤ x1 ≤ x0 WILL have a least element. Is that right?

Homework Statement Show that if X is a subset of a well-ordered set (A, ≤ ) such that x0 > x1 > x2... then X must be finite. The Attempt at a Solution It seems like there's an obvious solution in that we know X must be well ordered, so has a least element. But by the question, X has a...