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March 11th, 2019, 02:34 PM   #11
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It doesn't. The ordinal $\omega+1$ is infinite and has a last element.
Can you explain this in more detail? Seriously.
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March 11th, 2019, 03:11 PM   #12
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Can you explain this in more detail? Seriously.
Suppose you take the ordered set of natural numbers 0, 1, 2, 3, 4, ...; and after that sequence, you adjoin a symbol $\omega$. Then you have the ordered set 1, 2, 3, ..., $\omega$. This is a well-ordered set that is infinite and has a last element.

Another notation for that ordered set is $\omega + 1$. That's because $\omega$ is the symbol for the ordered set of natural numbers $\{0, 1, 2, 3, \dots \}$

You might ask, what exactly is $\omega$? In fact you can take it as the set 0, 1, 2, 3, ... itself. In other words $\omega + 1 = \{0, 1, 2, \dots, \{0, 1, 2, \dots\}\}$.

https://en.wikipedia.org/wiki/Ordinal_number
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Last edited by Maschke; March 11th, 2019 at 03:43 PM.
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March 11th, 2019, 03:58 PM   #13
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Originally Posted by Maschke View Post
Suppose you take the ordered set of natural numbers 0, 1, 2, 3, 4, ...; and after that sequence, you adjoin a symbol $\omega$. Then you have the ordered set 1, 2, 3, ..., $\omega$. This is a well-ordered set that is infinite and has a last element. Another notation for that ordered set is $\omega + 1$.

You might ask, what exactly is $\omega$? In fact you can take it as the set 0, 1, 2, 3, ... itself. In other words $\omega + 1 = \{0, 1, 2, \dots, \{0, 1, 2, \dots\}\}$.

https://en.wikipedia.org/wiki/Ordinal_number
I'm sure this is mathematical naivete but the words "and after that sequence" when applied to an infinite sequence throw me.
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March 11th, 2019, 04:06 PM   #14
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I'm sure this is mathematical naivete but the words "and after that sequence" when applied to an infinite sequence throw me.
We adjoin things to infinite sets all the time. For example if we have the real numbers we can adjoin a symbol $i$, give it the property that $i^2 = 1$, and then consider the set of all formal symbolic expressions $a + bi$ where $a$ and $b$ are real and the usual rules for multiplying binomials apply. That's one way to construct the complex numbers.

In this case we start with an ordered set $\{0, 1, 2, 3, \dots \}$ and we adjoin a new symbol $\omega$, with the rule that $n < \omega$ for any value of $n = 0, 1, 2, \dots$

This formal construction gives us a well-ordered set (every nonempty subset has a smallest member) but unlike the ordered set of natural numbers, our new augmented set has a largest element.

Here's another representation of the same idea. Start with the usual order on the natural numbers 1 < 2 < 3 < 4 < ... and make a new rule that n < 0 for all n = 1, 2, 3, ...

Now we still have the same underlying set of natural numbers $\{0, 1, 2, 3, \dots \}$ but with a new order: 1 < 2 < 3 < ... < 0.

In fact this is exactly the same order type as $\omega + 1$. It's the order type of the usual natural numbers with something stuck at the end. We haven't changed the set at all, it's still got all the same elements; but they're in a different order.

It might have been better if I'd just started with this example. Reorder <0, 1, 2, 3, ...> as <1, 2, 3, ..., 0> where the angle brackets indicate that these are ordered sets and not just sets. As sets they are the same, because they have the same elements. But the order is different; and in the second case, we have an infinite, well-ordered set that has a largest element.

Last edited by Maschke; March 11th, 2019 at 04:16 PM.
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March 12th, 2019, 07:26 AM   #15
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Don't let Zylo see that post.
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March 12th, 2019, 10:29 AM   #16
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This formal construction gives us a well-ordered set (every nonempty subset has a smallest member) but unlike the ordered set of natural numbers, our new augmented set has a largest element.
This reads to me like the definition of the hyperreal infinity.

Have they done the same thing with the natural numbers? Is there a hyperinteger infinity? Apparently they have.
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March 12th, 2019, 10:54 AM   #17
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This reads to me like the definition of the hyperreal infinity.

Have they done the same thing with the natural numbers? Is there a hyperinteger infinity? Apparently they have.
Nothing of the sort. I'm traveling today and haven't time to reply in detail, but please read the Wiki article on ordinals and see if it helps.

But look, I have the set {0, 1, 2, 3, ...} with the usual order 0 < 1 < 2 < 3 < ...

And now I have the EXACT SAME SET, so there are no new elements added, but with the "funny" order 1 '<' 2 '< 3 '<' ... '<' 0.

Same exact set in each case. In the usual order, no largest element. But in the funny order, there is a largest element. And both ordered sets are well-ordered by their respective orders, meaning that every nonempty subset has a smallest element.

It's like lining up a bunch of schoolkids by height, then asking them to line up in alpha order, then by test score, then by inverse alpha order. There are many ways of ordering a set. Of course in the finite case, all those distinct orders have the same order type; but in the infinite case, a reordering can change the order type.

Last edited by Maschke; March 12th, 2019 at 10:59 AM.
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March 12th, 2019, 04:49 PM   #18
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Thank you. I revisited some of my texts on set theory that hadn't seen the light of day in a while and I understand what you are talking about now.
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March 13th, 2019, 02:34 PM   #19
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There is no last digit to the expression for $\pi$.
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March 13th, 2019, 04:56 PM   #20
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There is no last digit to the expression for $\pi$.
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