Re: chains of subsets of N

On Thu, 25 Oct 2007 17:30:28 -0700, The World Wide Wade
<aderamey.addw@xxxxxxxxxxx> wrote:

In article <8d92i3lcbrgc4o50mqpn4e7lndp81u4sff@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

Two problems ...

I asked one of these in the thread "a recursive dilemma" but it went
unanswered.

Prove or disprove:

(1) An ascending chain of subsets of N cannot be uncountable.

Define an ascending chain of subsets of N please.

I'm assuming the subsets of N are partially ordered by inclusion.

But for clarity, I should have made that explicit.

(2) Every ascending chain of subsets of N can be extended to a maximal
chain.

I think (1) is true and (2) is false.

Not any more.

I now think that (1) is false and (2) is true.

In fact, assuming the axiom of choice, (2) _is_ true.

I think I can prove (1), but I'm not completely sure of my proof.

The proof of (1) vaporized as soon as I tried to pin it down on some
key details.

So for (1), I have neither a proof nor a disproof.

Also, I posed a 3rd problem in another reply.

quasi
.

Relevant Pages

  • Re: chains of subsets of N
    ... quasi wrote: ... Prove or disprove: ... An ascending chain of subsets of N cannot be uncountable. ... Does the argument above generalize to arbitrary cardinalities? ...
    (sci.math)
  • Re: chains of subsets of N
    ... quasi wrote: ... Prove or disprove: ... An ascending chain of subsets of N cannot be uncountable. ... In fact, assuming the axiom of choice, _is_ true. ...
    (sci.math)
  • Re: chains of subsets of N
    ... quasi wrote: ... Prove or disprove: ... There exists an ascending chain of subsets of N with no maximum ... just a momentary blind spot. ...
    (sci.math)
  • Re: chains of subsets of N
    ... quasi wrote: ... Prove or disprove: ... An ascending chain of subsets of N cannot be uncountable. ... a collection of sets S_a with a in some index set I ...
    (sci.math)
  • Re: chains of subsets of N
    ... Prove or disprove: ... An ascending chain of subsets of N cannot be uncountable. ... any chain can be extended to a maximal chain. ... I withdraw that claim as well. ...
    (sci.math)