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February 26th, 2018, 01:43 PM   #21
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Originally Posted by zylo View Post
I said Sn was not a natural number. I simply meant it to be an example of something constructed, or defined by, natural numbers. If it makes it easier, put spaces between.

S$\displaystyle _{11}$ = 1 2 3 4 5 6 7 8 9 10 11
$\displaystyle lim_{n \rightarrow \infty}$ Sn =1 2 3,......,12578, ..........


If none of {Sn} are unending, then there has to be a maximum value of n for Sn. There isn't.
Take Sn = n

S1 = 1, ending
S2 = 2, ending
S3 = 3, ending
S100 = 100, ending.

For which n is Sn non-ending?

Last edited by skipjack; February 27th, 2018 at 12:40 AM.
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February 26th, 2018, 01:59 PM   #22
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0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999.

P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............}

What is the largest member of {Pn}? I am not asking what any particular member of {Pn} is, I am asking what the largest member is.

Last edited by skipjack; February 27th, 2018 at 12:40 AM.
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February 26th, 2018, 01:59 PM   #23
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Originally Posted by Maschke View Post
Very interesting comment that I'd like to understand with your help. I dove into the hyperreals a couple of years ago. I read Terence Tao's articles on NSA including his awesome explanation of ultrafilters as voting systems. I know what a nonprincipal ultrafilter is. I read Lightstone's paper on his notation for hyperreals and I read a few other papers besides. There was a moment when I thought I had it all straight in my mind but apparently that clarity is gone now.

I mention this so that you can calibrate the state of my knowledge. Can you please explain how $(S_n)$ has a different limit than $(n)$ in the hyperreals or hyperintegers? Thanks much.
Ah well, perhaps some interesting discussion can come from this rubbish thread anyway.

So, the real numbers are often defined to be equivalence classes of rational Cauchy sequences. The hyperreal construction is very similar, but instead of taking Cauchy sequences, we take all sequences (incidentally, if we start with Q and do the hyperreal construction on it, then the result will contain R, it is an alternative construction to R!!)

But the set of all sequences isn't a field, indeed, what would the inverse of (1,0,1,0,1,0,...) be? To solve this, we introduce the ultrafilter which decides for us which elements to identify and which not (this is problematic since ultrafilters cannot be constructed at all, and different choices of ultrafilters give different hyperreals!!)

A real sequence $(x_n)_n$ would then have as hyperreal limit the exact same sequence. I find this very elegant. And it gives some surprising other results:
If we interpret 0.999999... as the limit of 0.9999...9, then this limit will be LESS than one, less by an infinitesimal amount.
The limit of the sequence (1,2,3,4,...) would be the hyperreal (1,2,3,4,...) and this hyperreal is different from (1, 12, 123, 1234,...) since their agreement set (there is only one place where it agrees, the first place) is finite, hence cannot lie in the ultrafilter.

Now, I have to be fair and say that this is not really how limits are defined in non-standard analysis. In non-standard analysis, we "fix" the issue that 0.999...=/=1 by basically saying that the limit operator should forget about the infinitesimal part. But to me the infinitesimal part is the nicest part, so I like my interpretation.

Last edited by skipjack; February 27th, 2018 at 12:43 AM.
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February 26th, 2018, 02:01 PM   #24
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Quote:
Originally Posted by zylo View Post
0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999.

P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............}

What is the largest member of {Pn}. I am not asking what any particular member of {Pn} is, I am asking what the largest member is.
There is no largest member, and $P\notin \{P_n\}$. Am I right?
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February 26th, 2018, 02:16 PM   #25
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Quote:
Originally Posted by zylo View Post
0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999.

P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............}

What is the largest member of {Pn}. I am not asking what any particular member of {Pn} is, I am asking what the largest member is.
So you have no response to my last post, above.

How did you manage to post your long post the instant afterI posted mine? Coincidence I guess.

"rubbish thread?" Can't do any better than that?

Last edited by zylo; February 26th, 2018 at 02:28 PM.
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February 26th, 2018, 02:53 PM   #26
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Alright, let's try a bit of a detour. Perhaps it has value, perhaps it doesn't.

We know that the rationals are countable. The reals consist of the rational numbers and the irrational numbers. The rationals are countable, and thus the irrationals must have the same cardinality as the reals because the irrationals can't be a finite set. (I hope we agree on that anyway.)

So let's construct the table using only the set of irrationals... we no longer need to consider the rationals and thankfully the issue of 0.9999999... = 1 is no longer an issue. For convenience, let's use the section (0, 1) of real numbers. Then our list to "count" the irrationals is of the form:
0.00222459875978...
0.39457935730945...
0.79874968734978...

etc.

We are back to Cantor's diagonal argument as the simplest way to work with the list, which I know that zylo will just looooove.

This table seems to me to be a cleaner argument than trying to take "limits" of real numbers.

-Dan

Last edited by skipjack; February 27th, 2018 at 12:50 AM.
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February 26th, 2018, 03:45 PM   #27
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Quote:
Originally Posted by zylo View Post
If none of {Sn} are unending, then there has to be a maximum value of n for Sn. Tkere isn't.
No there doesn't. $S_n$ terminates after $n$ elements. The infinite case $S$ doesn't terminate because it's infinite. If it terminated it would be finite.
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February 26th, 2018, 03:55 PM   #28
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Quote:
Originally Posted by topsquark View Post
Then our list to "count" the irrationals is of the form:
0.00222459875978...
0.39457935730945...
0.79874968734978...

etc.

We are back to Cantor's diagonal argument as the simplest way to work with the list,
I don't think this guarantees that your diagonal constructs an irrational. Indeed, it seems easy to guarantee that it doesn't: make every element $0.00\ldots012345678910111213\ldots$ with $n$ zeros at the start of the $n$th element.
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February 26th, 2018, 05:05 PM   #29
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Quote:
Originally Posted by zylo View Post
0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999.

P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............}

What is the largest member of {Pn}. I am not asking what any particular member of {Pn} is, I am asking what the largest member is.
Let $t_n = 1 - \frac1n$. Clearly $t = \lim \limits_{n \to \infty} t_n = 2$. So what is the largest member of the sequence $\{t_n\}$?

Clearly, there isn't one. All $t_n \gt t_{n-1}$ and all $t_n \lt 2$. So $t \not \in \{t_n\}$.
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February 26th, 2018, 07:14 PM   #30
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Originally Posted by zylo View Post
So you have no response to my last post, above.
I did respond. Look again.

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"rubbish thread?" Can't do any better than that?
Why would I do better? I'm not here to insult you, or disprove you or humiliate you. I'm just here to discuss interesting mathematics. And this thread has no interesting mathematics. And it's made worse by very, very confusing notations. Again: not meant to be insulting or disprove you.

So why reply then? Hey, man, it's 5 am where I live and I'm replying to a math thread. Why do you think? Obviously: I can't sleep and I'm bored.

Last edited by skipjack; February 27th, 2018 at 12:47 AM.
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