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From: |-|ercules on 27 Jun 2010 21:24

OK extending WM's example.

You come across a fork in a road with two options, left and right.

Next to the fork is a sign that reads
Each path left or right leads to a new fork with two more options, left or right.

Beside the sign is a shop, INFINITE PATH MAPS.

The shop contains infinite maps, each numbered 1, 2..

Each map is an infinite sequence of lefts and rights.

The shop is infinitely long, and each row has a sign that reads
All maps in this section contain all paths X levels deep (plus the remaining infinite paths of Lefts and Rights in each map)
The first row X=1, the second row X=2 and so on.

So the millionth row, contains well over 2^1,000,000 infinite maps that cover every path option
on the first million branches.

Considering if you have a map for a certain path you can always find your way back,
and you legs are solar powered bionics that never wear, and your cyborg body is suitably matched to your legs, the sun contains
infinite hydrogen fuel etc. etc.

would you get lost?

Herc

From: Transfer Principle on 28 Jun 2010 17:00

On Jun 27, 6:24 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> OK extending WM's example.
> You come across a fork in a road with two options, left and right.

Ah yes, WM's infinite binary tree example.

> Considering if you have a map for a certain path you can always find your way back,
> and you legs are solar powered bionics that never wear, and your cyborg body is suitably matched to your legs, the sun contains
> infinite hydrogen fuel etc. etc.
> would you get lost?

As I'm already familiar with WM's examples, I assume the point
of this is to show that according to the standard theory,
there are only countably many maps, yet there are uncountably
many infinitely paths. Thus anyone who believes in the
uncountability of the paths would be forced to answer "yes"
to the question "Would you get lost?" since there aren't
enough maps for all of the paths, and anyone who answers "no"
would be admitting that there are only countably many paths,
in accordance with Herc's and WM's beliefs.

I assume that the correct answer for someone who uses ZFC to
give would be something like, since each map is located only
a finite distance from the starting point (even though each
path extends away from the start infinitely), one can only
prove that there are only countably many initial _finite_
segments of each path, not that there are countably many
infinite paths.

Still, I see nothing wrong with having a theory which does
satisfy either Herc's or WM's intuitions.

From: Alan Smaill on 28 Jun 2010 17:41

Transfer Principle <lwalke3(a)lausd.net> writes:

> Still, I see nothing wrong with having a theory which does
> satisfy either Herc's or WM's intuitions.

Of course you're right;
there's nothing wrong with having such theories.

Is there anyone who has said anything different in sci.logic?

Do you think that Herc or WM has actually proposed
any such theory?

--
Alan Smaill

From: |-|ercules on 28 Jun 2010 19:04

"|-|ercules" <radgray123(a)yahoo.com> wrote
>
> XXXXX... is not a single real. It's shorthand for a Complete Digit Sequence
> for all widths.

I should explain my ambiguous use of variables.

In your counter examples, X is a standard variable for a real or a list.

In my demonstration, X is a 1 digit wide matrix, XX is a 2 digit wide matrix

Herc

From: Sylvia Else on 28 Jun 2010 20:08

On 28/06/2010 11:15 PM, |-|ercules wrote:

> You prove case n, by logical inspection, that the property holds for all
> elements in the list of size n. (a whole object)
>
> I prove case n, by logical inspection, that the property holds for
> prefixes of width n. (a subset of digits)
>
> You prove, by induction, that all (finite) sizes of lists the property
> holds.
>
> I prove, by induction, that the property holds for all digit widths.
> (all digits).

You only prove that it holds for finite digit widths. Somehow you want
to be allowed to extrapolate from finite to infinite when it suits you,
but not for me to allowed to do the same thing when it doesn't suit you.

>>
>> BTW, it occurs to me that if you have a list containing all the
>> prefixes obtained by permuting 1, 2, 3, ... n digits, followed in each
>> case by infinite suffixes, then we can use the anti-diagonal technique
>> to construct an infinite sequence A whose first k digits differ from
>> the corresponding diagonal digits of your list. That sequence A isn't
>> in your list.
>>
>> When you move from n digit prefixes to n+1 digit prefixes, you add
>> 10^n+1 new sequences, and we can change the next 10^n+1 digits of A so
>> that it is still not in the list.
>>
>> Thus at each induction step, you increase the number of prefixes in
>> the list, but you retain a sequence A that is not in the list. This
>> remains true for any finite number of steps.
>>
>> That is, no induction step extinguishes sequence A which is not in the
>> list. It is hardly reasonable to argue that extending this to an
>> infinite number of steps creates a list with every inifinite sequence,
>> but in the process mysteriously extinguishes sequence A.
>>
>> Sylvia.
>
> That's just a proof that finite lists are incomplete, and that sequences
> are shorter than
> the number of sequences to cover every digit sequence.
> Similar to the favorite argument of ANTI-Cantorians!
>
> A list contains all finite prefixes.
>
> Step 1: a 1 digit anti-diagonal is not a new sequence
> Step n: a n digit anti-diagonal is not a new sequence
> Step n+1: a n+1 digit anti-diagonal is not a new sequence (ok IF n is
> then n+1 is, but no base step!)

You're talking about finite length anti-diagonals. I clearly said
"construct an *infinite* sequence A whose *first* k digits differ from
the corresponding diagonal digits of your list."

As I said, A is not in your list, and it never will be no matter how far
you take the induction.

Sylvia.