March 11th, 2019, 01:34 PM  #11 
Senior Member Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276  
March 11th, 2019, 02:11 PM  #12 
Senior Member Joined: Aug 2012 Posts: 2,204 Thanks: 647  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 wellordered 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 Last edited by Maschke; March 11th, 2019 at 02:43 PM. 
March 11th, 2019, 02:58 PM  #13  
Senior Member Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276  Quote:
 
March 11th, 2019, 03:06 PM  #14  
Senior Member Joined: Aug 2012 Posts: 2,204 Thanks: 647  Quote:
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 wellordered 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, wellordered set that has a largest element. Last edited by Maschke; March 11th, 2019 at 03:16 PM.  
March 12th, 2019, 06:26 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,618 Thanks: 2608 Math Focus: Mainly analysis and algebra 
Don't let Zylo see that post.

March 12th, 2019, 09:29 AM  #16  
Senior Member Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276  Quote:
Have they done the same thing with the natural numbers? Is there a hyperinteger infinity? Apparently they have.  
March 12th, 2019, 09:54 AM  #17  
Senior Member Joined: Aug 2012 Posts: 2,204 Thanks: 647  Quote:
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 wellordered 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 09:59 AM.  
March 12th, 2019, 03:49 PM  #18 
Senior Member Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276 
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.

March 13th, 2019, 01:34 PM  #19 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
There is no last digit to the expression for $\pi$.

March 13th, 2019, 03:56 PM  #20 
Senior Member Joined: Oct 2009 Posts: 752 Thanks: 261  

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