Re: Countable unions of countable ordinals w/o choice


Bill Taylor wrote:
Rupert wrote:

it can be proved that there is a surjection from R to omega-1.
So if R is a countable union of countable sets, then so is omega-1

The conclusion may be true, but is this reasoning valid?

It seems to use the lemma that

exists surjection f:A --> B
implies
A is CUCS ==> B is CUCS.

But without choice how are you going to prove this?


Let A be the union of a countable family {A_n:n in N} of countable
sets. Let f be a surjection from A to B. Since A_n is countable, f(A_n)
will be countable for all n, this can be proved without choice. Since f
is surjective {f(A_n):n in N} will cover B.

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Bill Taylor W.Taylor@xxxxxxxxxxxxxxxxxxxxx
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Yesterday, I thought I had a new deja vu experience.
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