
Re: Uncountability of the Real Numbers Without Decimals
Posted:
Dec 2, 2013 10:02 PM


On Monday, December 2, 2013 2:23:01 PM UTC7, WM wrote: > Am Montag, 2. Dezember 2013 21:34:25 UTC+1 schrieb Zeit Geist: > > > > > The proof proceeds by Contradiction. We assume the Set of Real Numbers is Countable, and thus can be exhausted in a Sequence. > > > > And that is the point at which further reading is absolutely useless. > > > Then why did you keep reading? > > I did not. I know Cantor's first "proof" very well. >
If you didn't keep reading, how do you know it was a rendition of Cantor's First Proof?
> > > M_1 = empty set, M_{n+1} = (M_n U (n1, n])  {q_n} > > > > wher (n, n+1] is the semiopem interval that contains all rational numbers q with n < q =< n+1 and q_n is the nth rational in Cantor's enumeration of all rationals. > > > So, an enigmatic s of the rationals is possible? >
That was supposed to be, "... an enumeration of the rationals ... ". But I think you figured that out.
> > No. But if you think that it is possible, you can see it being contradicted here. >
Not so much.
> > > > The notion of countability requires that lim M_n = { } for n > oo. > > > No sober mind will accept that. > > > How do you know q_n is in your semiopen interval and isn't added in later? > > I start with the enumeration: 1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, ... The numbers q_n are always in the intervals already added. >
Then each rational is eventually counted at some time. Yes, at each step your Set is infinite. However, for some step, each rational is counted (removed). Thus, every rational is counted.
Strange how the idea of the infinite goes against our finite intuitions.
Also, strange how use an algorithm with infinite steps to refute the existence of infinite things. Although, your conclusion is False.
> > > > I find the proof rather straight forward. Question, comments, suggestions and corrections are welcome. > > > > What step produces the first undefinable number from a given set of defined numbers? > > > None or maybe any. The proof just uses existence nonempty intervals. Unless you think some of those intervals are empty? > > Your proof is only an implication, a frame. It shows you: If you apply a definable sequence, you will get a defined limit. Undefinable sequences will not produce anything. No mathematical operation will ever produce an undefinable number. Therefore it is useless to think they exist. They do not help to resolve Cantor's antinomy that his "proof" of uncountability is done by producing defined reals. (Do not think that you get an undefinable limit if you refrain from defining a sequence.) >
First, I have never brought up "definable" anything. This is your concept.
Second, if you bring refuse to bring "definable" into a Consistent Logical Mathematic conception, then I have no desire to discuss it with you.
> > And please discuss "definable reals" until you have solid Mathematical conception of the idea. > > That is not required for my purpose. It is completely sufficient that one finite word will at most define one number. It is not necessary to go into the details, since the superset of all finite definitions is countable  nothwithstanding how finite definitions are defined.
You are highly confused what countable means here.
Give me a bijection between { "definable real number" } and N. You will not be able to do so until you give a precise meaning of "definable".
> Look here: It is mathematical fact that all rationals that are less than 1 are less than 2. For that sake I need not answer the question what the largest rational less than 1 could be.
That makes no sense.
> But I understand that you have run out of solid arguments.
Talking to yourself?
My arguments may seem to be repeating because you objections are. And, you dont provide what I ask for, not that anything should compel you to. But, if your gonna argue something, defend it.
> > Regards, WM
ZG

