Re: Tea cups and elephants

On Nov 29, 6:36 pm, matt271829-n...@xxxxxxxxxxx wrote:
On Nov 29, 7:27 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:



On Nov 29, 9:34 am, matt271829-n...@xxxxxxxxxxx wrote:

On Nov 29, 2:00 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Nov 29, 12:56 am, matt271829-n...@xxxxxxxxxxx wrote:

On Nov 29, 4:36 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Nov 28, 8:19 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:

I have posed this question in one my recent threads as a response to
some posts, but with no answer.

Constraint: Every tea cup has exactly zero elephants in it's contents.

Theorem: The cumulative contents of an infinite number of tea cups
will have a finite number of elephants in them.

Can this theorem be proven while meeting the given Constraint?.

- venkat

Without additional assumptions the theorem cannot be shown to be true
or false.

With appropriate additional assumptions. E.g.

The number of elephants in a set of tea cups is
the sum of the number of elephants in each tea cup in the set.

The sum of any number of zeros is zero.

the theorem can be shown to be false.

Note that for sets of points the second assumption is true, but the
first
assumption is false. The extent of a set of points may not be the sum
of
the extents of each point in the set.

The fact that each point has extent 0 and an infinite number of
points may have finite extent, is a simple result of measure theory.

What about the plain old sum of plain old ordinary numerical zeros? Is
there any way that the sum of "infinitely many" zeros can be said to
be non-zero? The limit as n -> oo of n zeros is presumably zero,
but...?

I have always been kinda confused about this.

If you have a finite number of zeros then the sum is zero.

If you have a countably infinite number of zeros then the sum is zero
(this depends of course on you definitionof a countably infinite sum,
using the usual definiton, the sum is 0)

If you have an uncountable number of zeros then the sum is zero
(this depends of course on you definitionof an uncountably infinite
sum,
using the usual definiton, the sum is 0. Indeed the sum over any
uncountable set is only defined if the value at all but a countable
number of points is 0. So the uncountable sum is really the
countable sum)

Hence for all standard meanings of sum, the sum of zeros is zero.
Hence the statement

The sum of any number of zeros is zero

is true

I should have made it clear that I was referring to countable
infinity. What puzzles me about this problem are sums such as c/n + c/
n + c/n ... carried to n terms. Whatever the value of n, this sum
apparently equals c.

But if we let n -> oo then the individual terms can be made to differ
from zero by as small an amount as we like. Normally if something can
be made to differ from something else by as small an amount as we like
then we conclude that, in the limit, the two quantities are equal
(e.g. limit of 0.99999... is one). So it would seem here that, in the
limit, c/n equals zero, and thus the sum of "infinitely many" zeros is
c (i.e. anything you care to choose; i.e. undefined). Where did this
go wrong?

In assuming that if lim a*b = k then lim(a)*lim(b) = k

This may not be true if lim(a) = oo or lim(b)=oo. You have
given an example.

It is true that lim n->oo ((n)*(c/n)) = c. It is also true
that lim n->oo (n) = oo and lim n->oo (c/n) = 0.
However you cannot conclude that (lim n->oo n) * (lim n->oo c/n) = c.

OK, that seems clear. Is (lim n->oo n) * (lim n->oo c/n), being, if
you'll excuse it, "oo * 0", undefined then?

I wasn't intending to use multiplication though, so I don't think I
was directly assuming lim(a*b) = lim(a)*lim(b).

In the first case I have

lim n->oo (sum_i=1^n c/n) = c

Which hopefully is uncontroversial.

Indeed, Note that sum_i=1^n c/n = (n)*(c/n)

I'm having problems writing the
second case that represents the limit of summing the limits, which
leads to

0 + 0 + 0 + ... = 0

I want to write something like lim n->oo (sum_i=1^n (lim n->oo (c/n)))
= 0, but this doesn't seem right with the conflicting use of n.

Correct. you can only have one "lim n->oo"

I tried lim m->oo (sum_i=1^m (lim n->oo c/n)),


This is correct (note that this is (lim m->oo (m))*(lim n->oo c/n))


but having m and n
unrelated doesn't seem right either. Here we're evaluating the inner
limit first, and then summing, whereas I actually want the two limits
to be approached in tandem.

The two limits approach in tandem is lim n->oo (sum_i=1^n (c/n))


This is your problem. If the two limits are approached in tandem
then
you get one result. If the two limits are evaluated separately, you
get
another result. There is no reason to expect the results to be the
same.

- William Hughes

.

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