Ultrafilter Richness: Explore Options Beyond AC

  • Thread starter Thread starter BDV
  • Start date Start date
BDV
Messages
16
Reaction score
0
Hello,

With the axiom of choice, we are left with two options for ultrafilters:
a) principal ultrafilters, built from a singleton {x}.
b) nonpricipal filters of which all contain the cofinite filter, ergo complements of finite sets subalgebras.

Isn't this kind of flimsy? To get to more exotic/exciting objects does one:
give up AC
or
gives up the ultra in ultafilter (the A or X\A is in F condition)?
 
Physics news on Phys.org
Why flimsy? For example, using an ultrafilter construction and without abandoning AC, you end up with measurable cardinals, which can do all sorts of interesting things. What sort of "exotic" objects did you have in mind?
 
I meant more exotic than the principal ultrafilters and complements of finite set subalgebras.

I realized last evening that non-principal ultrafilters may also contain objects with complements greater than finite sets. So there may be some richness right there.

I just wanted to get a good feel of the conceptual reach/limits of ultrafilters before I sink a significant amount of work in them. Ars longa, vita brevis.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.

Similar threads

Replies
2
Views
3K
Replies
7
Views
3K
Replies
9
Views
3K
Back
Top