Cont: Deeper than primes - Continuation 2

[jsfisher]

What about: the product of all the numbers in an uncountable set exists if every sequence of multiplications of numbers in that set converges to the same number?

That does highlight the need for a definition of what product over an uncountable set would mean. Doron isn't at that step. He's not even at the step of product over a countably infinite set (which is defined). He still thinks he can extrapolate from the meaning of product of a pair of numbers.

(By the way, I assume you meant "every countably infinite sequence". Yes?)

It's zero though in this case.

If it exists, than, yes, it must be zero.

 

[caveman1917]

That does highlight the need for a definition of what product over an uncountable set would mean. Doron isn't at that step. He's not even at the step of product over a countably infinite set (which is defined). He still thinks he can extrapolate from the meaning of product of a pair of numbers.

His problem isn't at the level of being uncountable though, but at the level of being infinite. Rather than the unit interval you can consider set of rational numbers strictly between zero and one, the product of which would also be zero under this definition and hence fail the non-zero condition.

(By the way, I assume you meant "every countably infinite sequence". Yes?)

Yes, that is correct. I've made it a bit more precise in a later edit.

 

[caveman1917]

He's not even at the step of product over a countably infinite set (which is defined).

What is its definition?

 

[jsfisher]

What is its definition?

I was actually thinking of product over a countably infinite sequence; over a set is a bit trickier, isn't it. I'll retract my parenthetical remark since I don't know if there really is a standard definition for that. (The one you had offered for uncountable sets could be used for countable finite, too, couldn't it.)

 

[caveman1917]

I was actually thinking of product over a countably infinite sequence; over a set is a bit trickier, isn't it.

Yes.

(The one you had offered for uncountable sets could be used for countable finite, too, couldn't it.)

I think the following works[*] with any finite, countably infinite or uncountably infinite set S which has a topology and any commutative binary (n-ary?) operation R on S with identity element e:

Let T be a subset of S. Let U be the set of all sequences in T which converge on some t in T \ e which are either uncountably infinite if T is infinite or have cardinality equal to |T| otherwise. Then R(T) exists if there exists some s in S such that for each u in U the sequence of partials u_R converges to s.

* works as in generalizes multiplication and addition for the natural numbers, integers, rational numbers, real numbers and complex numbers and finite groups for which we can define limits. Though it's still not defined everywhere (for example on the interval [0, 2]) and is still ugly since you'd want the product of ]0, 1[ u {2} to be zero yet it wouldn't be defined.

ETA: nope, needs injection instead of just sequence, otherwise product of the set {2, 3} could be 2 * 2 instead of 2 * 3.

 

[doronshadmi]

I'm not a mathematician, but my reasoning is as follows:
IF the product exists, it must be greater than 0, since all the numbers that are multiplied are greater than zero.
However, if we call the product x, x can not be greater than zero, since if it was, infinitely many of the numbers that are multiplied would be smaller than x.

Therefore, the product can not exist.

How did I do?

x is not some particular real number, but it as a symbol of an ever smaller result of (0,1) members that are multiplied with each other, which prevents their accurate cardinality.

 

[doronshadmi]

Doron isn't at that step.

jsfisher, you are not in the step to realize that since 0 is not one of the multiplied numbers of (0,1) then:

1) No result is 0.

2) By multiplying all (0,1) members with each other, the result is always smaller than each one of them AND > 0.

3) Conclusion: The claim that there is a set of real numbers in (0,1) with an accurate cardinality, is false.

You also ignore http://math.stackexchange.com/questions/2105990/infinite-multiplicity.

 

[jsfisher]

1) No result is 0.

Were this a product over a finite set of numbers, then sure. It isn't.

2) By multiplying all (0,1) members with each other, the result is always smaller than each one of them AND > 0.

You still need to (a) prove such a product exists (which, as has been revealed, will also require a definition of what it means to take the product over an infinite set) and (b) prove such a product is greater than 0.

Simply repeating your claims accomplishes neither.

 

[jsfisher]

What about: the product of all the numbers in an uncountable set exists if every[*] sequence of multiplications of numbers in that set converges to the same number?



It's zero though in this case.

* ETA: actually it depends on the sequences you consider, so not every sequence but every sequence which converges. Differently, let S be an uncountably infinite subset of R. Let U be the set of sequences in S which converge to an element of S. Then the product of all numbers in S exists if there exists some r in R such that for every u in U the corresponding sequence of partial products converges to r. The product of all numbers in S is then equal to r.

Here's an interesting convergent sequence:

Start off with a₁ = (d₁-1) / d₁ = 3 / 4

Then let a_i+1 = (2*(d_i-1)-1) / (2*(d_i-1))

I believe the product over that sequence converges to 1/2.

 

[doronshadmi]

Were this a product over a finite set of numbers, then sure. It isn't.

... and (b) prove such a product is greater than 0.

All is needed is to show that 0 is not one of the multiplied numbers in (0,1), and this term is satisfied by the very definition of (0,1).

You still need to (a) prove such a product exists (which, as has been revealed, will also require a definition of what it means to take the product over an infinite set)

You are still missing the nature of such product that is not any particular (0,1) real number, as already very simply shown in http://www.internationalskeptics.com/forums/showpost.php?p=11676837&postcount=2326 (and we are talking there about not less than the infinitely many members of (0,1)).

Simply repeating your claims accomplishes neither.

Simply repeating on (a) (b) accomplishes no understanding of the issue at hand.

You are in a "good" company of can be clearly seen in http://math.stackexchange.com/questions/2105990/infinite-multiplicity.

 

[jsfisher]

All is needed is to show that 0 is not one of the multiplied numbers in (0,1), and this term is satisfied by the very definition of (0,1).

While true for finite sets, it is not true for infinite ones.

Be that as it may, you still lack a definition for what it means to take the product of the members of an infinite set, and then you need to prove the product over (0, 1) exists and the product is greater than zero.

So far, all you have done is assume.

 

[caveman1917]

Here's an interesting convergent sequence:

Start off with a₁ = (d₁-1) / d₁ = 3 / 4

Then let a_i+1 = (2*(d_i-1)-1) / (2*(d_i-1))

I believe the product over that sequence converges to 1/2.

For every x in ]0,1[ there exists a sequence in ]0,1[ the product over which converges to x. Hence my later edit in the post you quoted (note that if the product over the sequence converges to anything but 0 then the sequence itself converges to 1, hence why I disallowed sequences converging to the identity element in my later post).

Proof: Let x be in ]0,1[ and start off with a_1 > x. Let prod_i(a) be the partial product of the first i terms of a. Then let a_{i + 1} be such that prod_{i + 1}(a) = (x + prod_i(a)) / 2. Such an a_{i + 1} always exists since prod_{i + 1}(a) = prod_i(a) * a_{i + 1} < prod_i(a). Then prod converges to x.

Basically, pick a target element in the unit interval, start above it, and with each term go half of the remaining way towards the target element.

I'm giving up on this though, it just gets uglier by the minute with ever-more corner cases popping up. I'll go with "the product over a infinite set is undefined".

 

[doronshadmi]

While true for finite sets, it is not true for infinite ones.

It is true whether it is finite or infinite ones simply because 0 is not one of the multiplied numbers in case of (0,1) and b.t.w also in case of (0,1].

No extra information about the issue at hand is needed, unless one is closed under (a) (b) loop.

Be that as it may, you still lack a definition for what it means to take the product of the members of an infinite set, and then you need to prove the product over (0, 1) exists and the product is greater than zero.

So far, all you have done is assume.

So far, all you have done is (a) (b) loop.

 

[doronshadmi]

I'm giving up on this though, it just gets uglier by the minute with ever-more corner cases popping up. I'll go with "the product over a infinite set is undefined".

It just gets uglier and therefore you go with "the product over a infinite set is undefined" exactly because you are using the notion of partial products instead of simply understand that since 0 is not one of the multiplied numbers in (0,1) or (0,1], no result is equal to 0.

More details about my argument are already given in http://www.internationalskeptics.com/forums/showpost.php?p=11677266&postcount=2330, and they can't be understood as long as one uses the notion of partial products about the issue at hand.

 

[doronshadmi]

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=11621981&postcount=2307.

If no operation (where at least multiplication is a form of operation) is used among set's members, one must distinguish between finite or infinite sets.

If at least multiplication is used among set's members, one must not distinguish between finite or infinite sets.

 

[jsfisher]

It is true whether it is finite or infinite ones simply because 0 is not one of the multiplied numbers in case of (0,1) and b.t.w also in case of (0,1].

Wish it to be true all you like. It is still not.

The product over the sequence 1/2, 1/3, 1/4, 1/5,... is zero, yet zero does not ever appear as an multiplier.

 

[doronshadmi]

Wish it to be true all you like. It is still not.

The product over the sequence 1/2, 1/3, 1/4, 1/5,... is zero, yet zero does not ever appear as an multiplier.

Because you are using the notion of partial product, which is artificial.

The natural notion is to understand that since 0 is not one of the multiplied numbers the multiplication over 1/2, 1/3, 1/4, 1/5,... is not zero.

More details are seen in http://math.stackexchange.com/questions/2105990/infinite-multiplicity.

 

[jsfisher]

Because you are using the notion of partial product, which is artificial.

Not artificial at all. It is, however, defined. Whatever thing you have in mind isn't.

Let us know when you can define your product-over-an-infinite-set idea.

 

[doronshadmi]

Not artificial at all.

The traditional mathematical framework artificially deals with the finite and the infinite exactly because:

1) It does not distinguish between the finite and the infinite in case that Dedekind infinite is considered.

2) It does distinguish between the finite and the infinite in case that multiplication is considered.

 

[jsfisher]

The traditional mathematical framework artificially deals with the finite and the infinite exactly because:

1) It does not distinguish between the finite and the infinite in case that Dedekind infinite is considered.

2) It does distinguish between the finite and the infinite in case that multiplication is considered.

Yeah, ok, whatever.

Let me know when you can define what you mean by product over a set. If what you want to talk about isn't defined, then there isn't much point talking about it, now is there?