Issues with Ordinals

[sharnjilraedan]

Working in set theory, and I'm having issues with a couple of problems.

1) Given w is omega, the smallest ordinal, prove that w is not isomorphic to w + 1.

What I've done, is shown that since w is in w + 1, then w + 1 cannot be isomorphic to it, for it would be isomorphic to an initial segment of w + 1, a contradiction.

2) There exists 2Aleph

Comments

[notmrgarrison]

1) w is the smallest infinite ordinal/cardinal.
I'm not sure what isomorphic means here, I'm guessing there exists f s.t. for all a,b: a < b iff f(a) < f(b).

w+1 has a largest element, w doesn't.
Shouldn't be too hard from there.

3) I'm guessing N is the naturals without zero.
What f maps N to w isomorphically?

[sharnjilraedan]

Okay, it's Naturals with 0. (We defined them recursively, where 0 = empty set, then defined the successor function, thus 1 = S(0) = empty set U {empty set}, etcera).

I'm having issues troubles with 3, in the sense that finding said isomorphism that maps N x N where it preserves the lexicographic order.

[notmrgarrison]

Still not clear, I'm guessing N = {1, 2, 3, ...) and w = {0, 1, 2, ....}

Then f(a,b) = (a-1, b-1)

[sharnjilraedan]

Nope. N = {0, 1, 2, ...} and w = N

See, even if I go with the map f(a,b) = a*b, the problem is, we're looking at the multiplication of two elements in the range, rather than a pair of elements, and I'm having issues preserving the Lexicographic order in that.

Unless I'm defining w*w wrong.

[devitojason]

For 2, here's the idea:

Show that there are at most 2^aleph naught (abbreviate this as 2^a) well orderings and at least that many, then apply cantor-schroeder-bernstein theorem.

First, a well ordering is simply a REALLY nice subset of N x N, interpreted as a relation. Thus, there are at most P(NxN) = P(N) = 2^aleph naught well orderings.

Why are there at least as many?

Consider the usual ordering 0 < 1 < 2 < .... and let f:N-> N be any bijection.

Define a new ordering a< N without asking that that the f's be bijective, so that's an upperbound on the number of f

[sharnjilraedan]

Actually, I had started like this, but couldn't get the Less than point.

It's greatly appreciated the help!