Re: infinity

Tony Orlow wrote:
> Jonathan Hoyle said:
> > >> Correct, and we do this for each finite value (of which there are an
> > >> infinite number of them), each time resulting in a finite number.
> > >> That's what we are saying.
> > >
> > > Take another look. The first element, 1, is the sum of 1 1. The
> > > second, 2, is the sum of 2 1's. The third, 3, is the sum of, yes, that's
> > > right, 3 1's. Each is 1 greater than the last, which is a sum of a set
> > > of 1's, so each is a sum of 1 more 1 than its predecessor.
> >
> > That is mostly correct, although more properly listed: 0 is the first
> > element, 1 is the second, 2 is the third, etc. so that n-1 is the nth
> > element for any finite natural number n.
> One can start at 0 or 1. For the purposes of this argument, it is easier to
> start with 1.

Yes, right. Compared with the problems at the other "end" [ha ha: there
isn't another end, is there], there's no need to quibble about starting
at 1 or 0.

> > > Now, how many of these elements do you have? Omega?
> >
> > If by omega you mean the smallest infinite cardinal (usually referred
> > to as Aleph_0), then yes. (Omega is generally used when referring to
> > it as an ordinal, but this is merely a matter of convention.)
> So, you have omega elements, from the first to the omega-th.

Note that the more usual answer would be 'No', since 'Omega' isn't
usually the name of a set size. Your interlocutor is trying to be
helpful by reinterpreting 'omega' in this non-standard way, but I'm not
sure that's a good idea, since at some point in the future you'll be
able to quote this statement about 'omega' out of the context of its
reinterpretation. But anyway...

> > >> How many 1's are summed in the omega-th element?
> >
> > There is no omega-th element.
> I beg your pardon? Is omega an ordinal? Did you say there are omega elements?

A better way to say it (short of the best way, which is to say that the
cardinality of the set is aleph-0) might be to say that "TO-omega" is a
name for the size of this set.

> If you have 10 elements, can you NOT have a tenth?

Spare us the other examples. We are extremely familiar with the fact
that given a finite set, you can count it by singing a ditty, and using
the name where the ditty stops as the size of the set (and if the set
is a sequence of integers from 1, also the name of the last integer in
the set).

> How do you have some number of elements, and not have
> that number as an index of one of them?

Suppose the set of elements is not one that can be counted by singing a
ditty. Suppose the attempt to sing the ditty never stops - in that case
there cannot be a name in the ditty where it stops because it doesn't.
(I mean, do you _really_ deny this?)

"TO-omega" is certainly not in the ditty (even without the 'TO'), since
the ditty consists only of the elements 'one' 'two' 'three' 'four'
'five' 'six' 'seven' 'eight' 'nine' 'space'. Remember we can simplify
the names of decimal numbers by just reading them digit by digit, and
using 'space' to separate one number from the next one. So the ditty
certainly cannot stop at anything called 'omega', and 'omega' cannot be
the last element in the set (even if we stick our fingers in our ears
and pretend to think about the last element in a set that has no last
element).

> >
> > >> 1. 1 is a natural number.
> > >> 2. natural(n) -> natural(1+(n))
> > >> So, we generate the set:
> > >> 1
> > >> 1+(1)
> > >> 1+(1+(1))
> > >> 1+(1+(1+(1)))
> > >> 1+(1+(1+(1+(1))))
> > >> etc.
> > >>
> > >> This is the definition of your Peano numbers.
> >
> > Essentially, yes.
> >
> > >> I don't understand how you cannot see that generating
> > >> a successor using an increment an infinite number of
> > >> times yields an infinite element.
> >
> > Your inability to understand is the key here. This process never
> > creates an infinite element, only all the finite ones.
> Then it only has a finite number of successor operations, and a finite number
> of elements. That is my point.

"Point". Hmmph. It's not much of an argument, since it's obviously
false, despite your frantic repetition of it. Even if you want to say
it has "a finite number of elements", it's clear this does not mean
what anyone else means by "a finite number of elements". Do you think
that if these elements are counted against a ditty the ditty will
actually stop, at an actual finite number? No, of course you don't -
you accept that the finite number at which it stops is only tenuously
existing, cannot be "specified", and only "exists" in some vague sense
no-one else can understand. All right, call this set "TO-unboundedly
TO-finite".

> I have no inability to understand.

Oh yes you do. Meeeeuuurgh!


> > Sure I can, I just did it. Your inability to prove this proposition
> > should at least be a warning to you that your intuition is faulty. I
> > have a set with infinite range, yet every value is finite.
>
> No, you don't. You have a set with an infinite diameter, but that does not
> measure the actual range of values WITHIN the set. That is an outside measure
> of the set which is at least as big as the range, and sometimes bigger. There
> is no infinite difference in the set, so how can any difference in the set
> posibly be infinite?

It can't. Don't be so forgetful. We have told you hundreds of times
that there are no infinite differences in the set. No *single* element
is infinitely far from any other. But there is no bound on how far away
an arbitrary element can be. There cannot therefore be a single
(normal, proper, mathematically non-tenuously existent) element that
indicates the "range", because any element you choose has a neighbour
that is further away. You really, really, cannot grasp this, can you? I
bet Wolf Kirchmeir's grandchild would eventually explode with
frustration trying to explain it to you. Well, you are either engaged
on a practical joke of staggering pointlessness, or you are beyond
help.

<snip>

Brian Chandler
http://imaginatorium.org

.

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