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Topic: ? 533 Proof
Replies: 46   Last Post: Aug 4, 2014 8:39 PM

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Martin Shobe

Posts: 1,469
Registered: 3/11/12
Re: ? 533 Proof
Posted: Aug 4, 2014 4:02 PM
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On 8/4/2014 12:37 PM, mueckenh@rz.fh-augsburg.de wrote:
> On Monday, 4 August 2014 19:18:23 UTC+2, Martin Shobe wrote:
>> On 8/4/2014 8:47 AM, mueckenh@rz.fh-augsburg.de wrote:
>>> On Monday, 4 August 2014 15:22:24 UTC+2, Martin Shobe wrote:
>>>>> If my proof concerning all finite initial segments is not accepted by you as sufficient, then you must believe that there is something in |N that isn't in at least one initial segment. Or have you a tertium datur?

>>>> I noticed you didn't answer the question. Anyway, there is no need to
>>>> believe that there is something in N that isn't in at least one finite
>>>> initial segment just because I found a flaw in your "proof".


>>> So, did you? Interesting. Excuse me when don't believe it, before you have shown the "flaw".

>> I have shown the flaw. What's your excuse for not believing it now?

> The lemma of my proof holds for every n that is mapped on an a/1. That was intended and makes the theorem true. In fact the theorem is true for every n. The proof could be reformulated to make also the lemma true for every n. But why should it? There is every k in |N of not enumerated rationals.

You're babbling. Try again, but make it coherent this time.

>>> But more interesting is this: Without that purported "flaw" there was obviously the need to believe that there is something in N that isn't in at least one finite initial segment.

>> There is no such need.

> You said: there is no need to believe that there is something in N that isn't in at least one finite initial segment just because I found a flaw in your "proof".

> The "because" is difficult to understand. The "flaw" concerns only a lemma. The theorem is correct. So why "because"?

You said, "If my proof concerning all finite initial segments is not
accepted by you as sufficient, then you must believe that there is
something in |N that isn't in at least one initial segment. Or have you
a tertium datur?"

Your proof is not sufficient since I found a flaw in it. That does not
mean that I have to believe that there is something in N that isn't in
at least one finite initial segment.

Martin Shobe




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