Re: set theory question

On Sat, 24 Jan 2009 03:12:54 -0500, quasi <quasi@xxxxxxxx> wrote:

On Fri, 23 Jan 2009 21:47:22 -0800, William Elliot
<marsh@xxxxxxxxxxxxxxxx> wrote:

On Fri, 23 Jan 2009, quasi wrote:
<tomtim20@xxxxxxxxx> wrote:

Mr. tomtim20 doesn't write. He scribbles.

let A is subset of R, and A is well ordered by < (usual order). prove
that A is countable.

A must have a least element -- call it a_1.

A \ {x_1} must have a least element, call it a_2.

etc.

Is the sequence a_1, a_2, a_3, ... monotonic?

Consider the open intervals between successive terms of the sequence
a_1, a_2, a_3, ...

Can there be uncountably many such intervals?

That seems elaborately vague.

Hints are allowed to be vague. If anything, I said too much.

But actually, now that I think about it, my hints don't really suffice
since my construction doesn't necessarily exhaust A. There could be
limit points. Thus, the simple plan I suggested doesn't actually work
(it was too simple). Sorry for having previously snipped your (William
Elliot's) suggested strategy (I didn't read it the first time), but
I'll put it back:

Assume A is uncountable. Thus
omega_1 order embeds in R. From that directly construct a
collection of nonnul pairwise disjoint intervals.

To William Elliot:

As noted above, I missed your point on first reading (sorry), but I'm
glad I reread it since it made me realize that the plan I suggested
was badly flawed.

To the OP:

Here's a set of replacement hints ...

(1) For each a in A, let U(a) be an open interval defined as follows:

If A has a largest element a_max, let U(a_max) = (a_max, oo)

For any other element a of A, let U(a) = (a,a')
where a' is the smallest element of A which exceeds a.

(2) Argue that the intervals U(a) are pairwise disjoint.

(3) A set of pairwise disjoint open intervals must be countable. Why?

(4) Is the function U one-to-one?

(5) Conclude that A is countable.

quasi
.

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