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 May 17th, 2017, 04:26 PM #1 Newbie   Joined: Apr 2017 From: Neither here nor there Posts: 3 Thanks: 0 Ordinal Arithmetic Problem Statement: Show that the set X of all ordinals less than the first uncountable ordinal is countably compact but not compact. Let μ be the first uncountable ordinal. The latter question is easy to show, but I stumbled upon a curiosity while attempting the former. In showing the former, I simply tried to show that every infinite subset of X should have a limit point (or in particular, an ω-accumulation point) in X. And so, in doing this, I needed to ensure that any infinite subset with μ as a limit point has another limit point in X. I reasoned that the first ω ordinals of this subset should only span a countable range of ordinals, since each of their co-initials are countable and a countable union of countable sets is countable. Any neighborhood of μ, however, is uncountable, so the limit point of the first ω ordinals of this subset cannot be μ. But when I considered the following set - The sequence {$ω^n$} = $ω$, $ω^2$, ... , $ω^n$, ... - it was hard to discern a limit point other than $ω^ω$. Aside from what the exact nature of the first uncountable ordinal is chosen to be, there should still be a limit point somewhere before $ω^ω$. So simply put, what ordinals exist between $ω^ω$ and the sequence I presented?
 August 7th, 2017, 10:37 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,050 Thanks: 1395 Do you still need help with this?

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