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Topic: Matheology § 133
Replies: 15   Last Post: Nov 6, 2012 10:37 AM

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rt servo

Posts: 19
Registered: 2/1/05
Re: Matheology § 133
Posted: Nov 1, 2012 3:19 PM
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On 01/11/2012 3:32 PM, MoeBlee wrote:
> On Nov 1, 1:10 pm, Alan Smaill <> wrote:
>> MoeBlee <> writes:
>>> On Nov 1, 12:40 pm, Alan Smaill <> wrote:
>>>> I don't think that one works --
>>>> well, not as diagonalising out of the reals, as opposed to out of
>>>> sequences of 0s and 1s. Multiple representations get in the way.

>>> Yes, I mean denumerable binary sequences as representations of members
>>> of the real interval [0 1]. I hoped that would have been clear from
>>> context?

>> OK, but in that case you do not know that the anti-diagonal real
>> is not present in the initial listing of reals, which is what
>> you surely really want. For all we know, it may be present
>> in a different representation.

> Maybe I'm foggy today. But is my recollection incorrect that, using
> the standard base-2 scheme, each member of [0 1] is represented
> UNIQUELY by a denumerable binary sequence. (What member of [0 1] is
> represented by more than one denumerable binary sequence in the
> standard representational scheme? 0 is represented by the sequence
> whose range is {0}, i.e, .0000..., and 1 is represented by the
> sequence whose range is {1}, i.e., in base 2 representation, .111...,
> right? So what member of [0 1] is represented by more than one
> denumerable binary sequence?)

Well, 0.1000... = 0.0111..., for starters. (Unless you take
'denumerable' to mean "ending in infinitely many 1s'.)

> And wasn't that the approach used by Cantor himself? Doesn't he take
> an enumeration of denumerable binary sequences and then show a
> denumerable binary sequence not in the enumeration?

I don't think he dealt with multiple representations in his proof. He
talked about infinite binary sequences, made up of symbols 'm' and 'w',
but as far as I remember, he did not interpret them as binary
representations of real numbers.

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