20170601, 19:47  #1 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Heights of Models of PA
This question is inspired by an incomplete solution of mine to a final exam problem a few years back.
For a countable model of , let Then  with  By Compactness + Downward LoweinheimSkolem, for every there is with Questions  Can we get ?  Can we get for arbitrarily large countable ?  Anything else interesting to say about ? 
20170602, 08:01  #2 
Dec 2012
The Netherlands
5×331 Posts 
Does PA stand for Peano Arithmetic here?

20170602, 19:41  #3 
Dec 2003
Hopefully Near M48
3336_{8} Posts 
Yup

20170603, 13:03  #4  
Romulan Interpreter
Jun 2011
Thailand
3^{3}×347 Posts 
Quote:
2. is there anything "in between"? I don't know about that. 3. yes, it is cute... Some guy like Cantor or Frankel may be able to tell more... hehe. For sure not me... Last fiddled with by LaurV on 20170603 at 13:05 

20170603, 16:15  #5 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2^{2}×3×887 Posts 
Which would you prefer to have? Choose one and run with it.
The "continuum hypothesis", that there is nothing in between, has been shown to be undecidable in ZFC. You can add it as an axiom, if you wish, or you can add its negation and each choice will lead to a consistent system. Last fiddled with by xilman on 20170603 at 16:15 Reason: Fix tag 
20170603, 16:48  #6 
Romulan Interpreter
Jun 2011
Thailand
3^{3}×347 Posts 
Hmm.. we have to google for this and read the "news" about... First, we don't know the English terms, and then we have a lot of lacunes (why is this red? is it lacunas? or lacunae?) in the math itself. We were once good at these things (set theory) but we feel like few milenia passed since...

20170603, 20:07  #7 
Dec 2003
Hopefully Near M48
3336_{8} Posts 
1) Well (the set of rationals) has height , despite being countable. One can show by transfinite induction that every countable embeds into , and taking the of all such yields .
The only problem with this example? It's not a model of . 2) No question about here, as this question is about . Last fiddled with by jinydu on 20170603 at 20:09 
20170604, 02:13  #8 
Romulan Interpreter
Jun 2011
Thailand
3^{3}×347 Posts 
That is new for me, and I think is false. According with my (old) memory, countable means \(\aleph_0\). There is a bijection between Q and N, and I still remember my high school teacher, Mrs. Diaconu, paining that diagonalcounting matrix on the blackboard.

20170604, 07:13  #9  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:


20170604, 09:57  #10 
Romulan Interpreter
Jun 2011
Thailand
3^{3}·347 Posts 
Well, the height, as I understand it, is the cardinal of the longest chain that preserves the order. I may be totally wrong here, but I can't see how I can make a chain in Q that preserves the order and yet, have more elements than Q itself. What I am missing?

20170604, 12:33  #11 
Jan 2017
2^{3}·11 Posts 
Height (as defined in the first post in this thread) is the sup of all the embeddable ordinals. You can't embed a larger cardinality, but you can embed all countable ordinals. Thus the sup is the first ordinal that is NOT countable.
