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Topic: WM's argument disassembled - If you are his student, please PLEASE comment here
Replies: 4   Last Post: Jul 24, 2014 5:41 PM

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 Ben Bacarisse Posts: 1,972 Registered: 7/4/07
Re: WM's argument disassembled - If you are his student, please PLEASE comment here
Posted: Jul 24, 2014 10:53 AM

mueckenh@rz.fh-augsburg.de writes:

> On Wednesday, 23 July 2014 23:35:48 UTC+2, Ben Bacarisse wrote:
>
>

>> From you correction here, I now see what you mean. lim |M_k|, where M_k
>> is the set {1/n | n < k} is indeed improper.

>
> So we agree.
>

>> Goodness know what you
>> think this shows but, yes, it is.

>
> This shows that there is no proper limit, namely a value that is
> approached as closely as desired. Sequences of sets cannot approach
> limits as closesly as desired, because the elements are quantized,
> i.e., not continuous. The smallest epsilon is one element. The only
> exeption is a constant sequence. Every term is the limit.
>

>> >> Obviously I can not interpret this since the term "improper limits" does
>> >> not mean the same to both of us.

>
>> > It means the same to everybody. The limit is infinite.
>>
>> But here you are talking about a set limit. The term improper applies
>> only to numerical limits. What is an improper set limit?

>
> That is a set that is not approached better than every positive
> difference. Since the terms of sequences of sets (if notbeing
> constant) differ by at least one element from step to step, the
> minimum eps is an element. But that is not "as small as desired".
>

>> Anyway, none of the limits involved in defining the set limits that I
>> wrote about are improper limits.

>
> Every set limit is an improper limit, unless the sequence is constant.

That's your own invention. The term does not mean that to anyone else.
Note that you did not address my point at all. I said that the limits
involved in defining the limit set are all proper. Since you said
nothing about these numerical limits, can I take it that you agree --
all the numerical limits used in defining the limit sets in question are
proper limits?

>> | Next, look at the examples in the paper I wrote for your students.
>> | Which of the limits there do you reject as needing completed
>> | infinity
>>
>> If you want to ignore some question or other, just say so.

>
> I answer herewith again: Every sequence of sets that is not constant
> (from some n_0 on) has no proper limit.

Your only answer is an assertion that relies on a term you only just
made up? Note that this newly invented term does not rely on completed
infinity, so even if it made sense it could not be an answer to my
question.

In what way does the set limit depend on completed infinity? What is
wrong with any of the ordinary numerical limits that are used to define
it?

>> > A proper limit does never depend on finished infinity.
>
>> Is that a yes? Does the set limit lim X_n where X_n = {1} for every n
>>
>> depend on either an improper limit or on completed infinity?

>
> This is a proper limit since for every element, the difference to the
> limit is less than any positive difference.

Great, so at least one set sequence has a limit. That's real progress.
You can't say now that they don't, only that you reject some, or most.

What about a set sequence that is constant for n greater than some
constant m? For example

s_n = {1..n} for n < 100
s_n = {1} for n >= 100?

Does lim s_n exists and is it {1}?

>> > I have explained what a proper limit is. I have shown that the limit
>> > of the cardinalities |s_n| in my example is an improper limit and that
>> > the improper limit s_n = { }
>> > has only the meaning that you advocated:

>>
>> > It is not a final set s_omega but is simply expressing that up to
>> > every n every rational number q_n will get enumerated without saying
>> > anything about all rational numbers.

>>
>> No, that's your own words. The limit tells you something about all
>> rationals (or every rational -- its the same in mathematics).

>
> So you believe in "all".

<Sigh>. Of course. It has never been anything but your fantasy that I
don't. I don't believe what *you* are saying about "completed
infinity", but that has no effect on what *I* think about infinite
sets.

For example, this nonsense that the limit set depends on "completed
infinity" is obvious bunk, even to you. Why else would you now say the
limits are "improper" for another reason altogether? It's because you
can't explain how "completed infinity" has anything to do with it, so

> What is the meaning of lim s_n = { }?

How many times do you need me to say it? I spelled it out in the paper
I wrote. It means that the limit of every indicator function is zero
(and see below for another way to put that).

> It is obviously not the meaning
> of calculus that the sets s_n come closer and closer to the empty
> set. They all are infinite. Is the meaning "a final state" s_omega
> approached with no regard to any finite state s_n?

No. It means (in this case) that every q_i is in only finitely many s_n
(the general meaning needs another clause, but let's keep it simple --
Virgil has given it dozens of times so you can't say you don't know the
full meaning).

> Then you should explain what makes the sets of the sequence empty "in
> the final state".

The definition does -- the same thing that "makes" 11 prime.

If, in your world, something "makes" lim 1/n = 0, then apply that to the
indicator functions that define the set limits. Whatever "makes" one
limit "makes" the other.

--
Ben.

Date Subject Author
7/24/14 Ben Bacarisse
7/24/14 mueckenh@rz.fh-augsburg.de
7/24/14 Tucsondrew@me.com
7/24/14 Virgil