Re: just how many reals are there in aleph_1 ?

From: Mark Nudelman (markn_at_greenwoodsoftware.com)
Date: 02/04/05 

Date: Thu, 3 Feb 2005 22:41:05 -0800

|-|erc wrote:
> No dumbass, the total of numbers on the blackboard is 21.
> That's a number and its well defined.

I have no desire to participate in a discussion with someone who uses
epithets like this. This will be my last posting in this thread.

You're also wrong. After you write 21 on the board, the total is now 42. I
think you missed the point of your own puzzle.

> 1 > The blackboard can contain any possible number.
> 2 > the total of numbers on blackboard is number.
> 3 > So it must be possible to put the total on the blackboard
>>> somewhere.
>>>
>>> Which step do you disagree with?
>
>
> So that's step 2 you disagree with is it? Do you realise
> when a question has 3 options your answer should begin
> with one of those options, followed by the explanation?

I thought my response was clear enough, but I guess you had trouble with it.

> You can't DEFINE a number to be off a complete list!

It's NOT a "complete" list (if by that you mean a list that contains all
reals). You're only assuming (incorrectly as it turns out) that the list is
complete. If you want to claim the list is complete, you must prove it, not
just assert it. But you'd be wasting your time, since the list is NOT
complete. And you can't DEFINE a list to be complete, without knowing
whether it's possible for such a thing to exist.

Look, here's a similar proof that sqrt(2) is rational:
Define a and b to be integers such that a^2 = 2*b^2. Now with some simple
manipulations, I prove that sqrt(2) = a/b. QED.
You see? I "defined" an impossible object (a and b), and ended up with a
false conclusion.

>> Ok, good, let's take a closer look at this. Let R[i] be the i-th
>> number on the list. Let d[i,j] be the j-th digit of R[i]. Now I
>> define Rx as the antidiagonal in the usual way -- Rx is the number
>> whose digits, which I'll call dx[j], are defined as
>> dx[j] = d[j,j]+1 mod 10, for all j
>> Is this "legitimately specified"?
>
> no. Silver Screen p 352 (just randomly flicked there, I'm Jesus
> remember)

[totally irrelevant quote deleted]

>> Now you're saying that
>> (a) Rx _is_ on the list, at some definite position k. That is Rx =
>> R[k]. Right so far?
>>
>> For two reals to be equal, each of their digits must be equal
>> (ignoring the infinite 9s issue). So since Rx = R[k],
>> (b) dx[j] = d[k,j], for all j. Right?
>>
>> But by the definition of Rx, dx[k] = d[k,k]+1, so
>> (c) dx[k] =/= d[k,k]
>> So (b) is not true (specifically, it's not true when j=k).
>>
>> So the assumption that Rx is on the list leads to a contradition --
>> (c) contradicts (b). The usual conclusion is that the assumption
>> was wrong, and Rx is not on the list. Do you have some other way to
>> resolve this contradiction?
>
> If the list is complete, then any number you define must be in it,
> with that assumption.

True, but the list is NOT complete. I showed a proof that it isn't
complete. You have yet to show any proof that it is.

> i.e. R[x] is on R is the assumption, for every R[x]

You can't ASSUME that R is a complete list, since what we're trying to
determine is whether such a thing as a complete list exists. If you start
by assuming that R has an impossible property, then use it to prove that R
indeed does have that property, you haven't proved anything.

> Then your antidiag is
> dx[j] = d[j,j] + 1 for all j.
>
> when j=x
>
> d[j,j] = d[j,j] + 1 , since the assumption is all numbers are on
> the list
>
> your definition is a contradiction

The definition of Rx is a contradition? In other words, you're saying it's
impossible to construct the anti-diagonal. I simply cannot create a number
whose digits differ from the digits in the diagonal. Right.

Goodbye.

--Mark

Relevant Pages

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