Re: An uncountable countable set


Tony Orlow wrote:
stephen@xxxxxxxxxx wrote:
imaginatorium@xxxxxxxxxxxxx wrote:

stephen@xxxxxxxxxx wrote:
imaginatorium@xxxxxxxxxxxxx wrote:
stephen@xxxxxxxxxx wrote:
Tony's argument seems to be a upside-down inside-out version
of one of Zeno's paradoxes.

...111111 (in binary) = 1 + 2 + 4 + 8 + ...

All the numbers on the right are finite, so the sum must
be finite. Apparently a sum can only be infinite if one of the
terms in the sum is infinite. An infinite sum of finite terms
cannot possibly be infinite in Tony's mind.
Not entirely sure about this. The problem above is that we are
restricting the string
...11111
so that every 1 is in a finite position. The position numbers never
achieve true infinity.
Yes, which means that each term in the sum
1 + 2 + 4 + 8 + ...
never achieves true infinity. Apparently because no single
term in the sum is infinite, the sum itself cannot be infinite.
If you had a 1 in an infinite position p, then the sum would
contain the term 2^p, which would be infinite, because p
is infinite, and the sum would be infinite.

It certainly sounds like the inductive proof that no natural is
infinite, since each is only 1 greater than the last, and incrementing a
finite always results in a finite. That same argument structure could be
applied to adding 2^n, which is always finite, producing a finite sum.
Since this applies to each and every bit position indexed with a finite
natural, it would seem it applies to each and every location within the
string, no? There is no possible point in the string where it can become
infinite.

Two statements :

1) A finite string of 1s represents a (finite) natural number.
2) An infinite string of 1s represents a (finite) natural number.

1) doesn't imply 2).



Hmm, I don't think you're quite getting into the spirit of this.

I am trying to figure out what Tony's misconceptions are.
Occassionaly I just make fun of them, but at other times
I try to see if I can figure out his intuition.

Wolfgang and I see the same problem. You start with 0 and start
enumerating the naturals using increment to generate the next successor.
Increment is defined as the addition of 1. So, the first is 1, the
second is 2, etc, etc. Where you have a set of consecutive naturals
starting at 1, the nth is n, the sum of n increments, and the size of
the set is always equal to the greatest value in the set.

What, even if there isn't a greatest value in the set?

So, if you
claim the size of the set is aleph_0, then that is also the greatest
natural, and is therefore finite by definition. Wolfgang and I object to
this being called infinite, since the identity relation between element
count and value in the naturals makes it impossible for the set to be
infinite without containing any infinite values.

Babble.

In
Tony's scheme (insofar as I understand it, and of course it appears to
me to be as contradictory as it does to you), I believe you can count
numbers 1, 2, 3, ... and so on, and go on for ever.

Yes, I have the same understanding of Tony's scheme.

Do you have a different scheme?

Standard theory? What maths were you hoping to do with your new scheme
that standard theory can't do, again?

But at some foggy
point (perhaps after about ever has elapsed) you find that you have
arrived in the infinite zone, and are counting numbers which, in an
infinite binary notation, have nonzero digits at least close to the
(nonexistent) left end. One of these numbers is called Big'un, and if
you write the sum

1 + 2 + 4 + 8 + ... + 2^Big'un + 2^(Big'un+1) + ...

then it _is_ infinite (in Tony's scheme), because at least one of the
numbers in the sum has achieved genuine infinity.

This seems to be the same understanding I have.

It's the difference between potential and actual infinity. Given N=S^L,
the only way to have an actually infinite number of strings in your
number system (N) is either to have strings of actually infinite length
(L), or a set of symbols (S) which is actually infinite. Since you have
a finite alphabet (0 and 1) and only allow finite strings, you cannot
have an infinite set of strings in the set.

So... there are a finite number of strings of 1s? So I should be able
to list them all in a finite amount of time...

1
11
111
1111
11111
111111

Wait, before I get into this, when am I going to finish?

1111111
11111111
111111111
1111111111

are you *sure* there a finite number of these things? How long till I
get to the end?

So in Tonyspeak, you have to quantify 1+2+4+... by saying what sort of
zone the dots extend to.

Precisely. You need to state a value range over which sets will be
compared, as a variable which can assume infinite values like Big'un.

Define BigUn non-circularly.

In this particular case we are talking only about the finite values.
So 1+2+4+... for all the finite values. According to Tony this
must be finite, because the sum of finite values, even an endless
sum of finite values, must be finite.

Since the "endless" sequence is unbounded but only potentially infinite,
the sum is potentially, but not actually, infinite. If you are allowed
to have an infinite iteration in that sum, say step #Big'un, you will be
adding an infinite value to the sum, 2^Big'un. So, they go hand in hand,
the count and the value of the term in the sum.

Babble.

I think.

The point of my post is that Tony's intuition is sort of the
opposite of the intuition that makes Zeno's paradoxes so compelling.
The idea that an infinite number of finite numbers can "sum" to
a finite value is a bit counterintuitive. The idea that you can
keep adding to something, and that it keeps getting bigger, but
that it remains finite, has puzzled (some) people for millenia.
Tony has twisted this around, and insists that an endless (infinite)
sum of finite numbers must be finite.

If no term has an infinite number of predecessors, then I don't consider
it an actually infinite sum.

Odd. Most people consider it an infinite sum if every term has an
infinite number of successors. How backwards.

Tony's intuition is based on the idea that numbers are processes.
...11111111 = 1 + 2 + 4 + 8 + 16 + ... (all finite powers of 2)
If I add the numbers on the right one at a time, I will always
have a finite number: 1, 3, 7, 15, 31, ... Tony therefore
concludes that the right hand side must be finite. There
is no step at which it "becomes" infinite. This is true.

Yes, there is no bit in that string where the sum becomes infinite.

So?

Two statements :

1) A finite string of 1s represents a (finite) natural number.
2) An infinite string of 1s represents a (finite) natural number.

1) doesn't imply 2).


Neither is there a step at which I have added all the numbers
on the right. ...11111111 equals the sum of all the numbers
on the right. So according to Tony's logic (if he was consistent),
there is no step at which the sum on the right becomes the value
on the left. So either the two are not equal, in which case
the fact that the right hand side is "finite" tells us nothing
about the value on the left. Or the right hand side "becomes"
infinite on the same step that it "becomes equal" to the left
hand side.

Stephen


It is not possible to identify any last step, that's true, but
irrelevant. I am not hinging my argument on any "last natural" as you
are trying to do. I am saying that for every bit position this fact
holds, and since that's all there is in the string, the fact holds for
the full extent of the string.

OK, for every "bit positon" the following fact holds :

"I could write the string up to that bit position down by hand, given
enough time."

Does this hold for the full extent of the string?

There is no need to talk about "getting
to the end" of that string, if it is inductively provable, as it is,
that no bit can exist in the string where an infinite value has been
achieved. You must go farther than any finite number of bits to get to
an infinite value, in which case you have farther than any finite number
of steps to get back to the first.

Tony

So... in conclusion you still don't know the difference between the
elements of a set or sequence and the set or sequence itself??

--
mike.

.

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