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Re: G_delta
Posted:
Jan 15, 2013 5:01 AM
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On Jan 15, 3:22 am, William Elliot <ma...@panix.com> wrote:
> On Mon, 14 Jan 2013, Butch Malahide wrote:
> > On Jan 14, 11:10 pm, William Elliot <ma...@panix.com> wrote:
> > > Does this generalize to every uncountable limit ordinal eta,
> > > that f in C(eta,R) is eventually constant and thusly the Cech
> > > Stone compactification of of eta is eta + 1? Does eta need
> > > to have an uncountable cofinality for this generalization?
>
> > I'd expect cofinality to have a lot to do with it, wouldn't you?
>
> Indeed, f in C(omega_1 + omega_0, R) is not eventually constant.
>
> If eta has finite cofinality, then any f in C(eta,R) is eventaully
> constant, namely from the last element of eta onward. That however
> isn't the full story for if f in C(omega_1 + 1, R), then f is
> constant from some xi < omega_1 and not just from the last element.
>
> If eta has denumberable cofinality, does this work to show
> f in C(eta,R) is eventually constant?
What does "denumberable" mean? Is that a real word, or did you make it
up? Does omega_1 + omega_0 have "denumberable cofinality"?
> Let (aj)_j be an increaing sequence within eta with denumberable
> cofinality.
Just any old increasing sequence within eta? You don't care if it
converges to eta or not? In that case why not simply set aj = j?
> Let K = { aj + 1 | j in N }.
> Then f(eta\K) = {0}, f(aj) = j, j in N is in C(eta,R)
> and isn't eventually constant.
Is this supposed to be the answer to the question you asked in the
previous paragraph? Why did you ask the question if you knew the
answer?
However, your function f is not well-defined. What is f(a1)? On the
one hand, you say that f(aj) = j for j in N, so f(a1) = 1. On the
other hand, f(a1) = 0 since a1 is an element of the set eta\K.
> What happens went the cofinality of eta is uncountable?
> Is f in C(eta,R) eventually constant?
Seems plausible enough. Are you saying that the method of proof used
for omega_1 doesn't work for ordinals of uncountable cardinality?
Where does it break down?
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