MHB Unbounded subset of ordinals a set?

  • Thread starter Thread starter RWood
  • Start date Start date
  • Tags Tags
    Set
AI Thread Summary
An unbounded subset C of the class of all ordinals R cannot be a set, but rather a class. This conclusion arises from the ability to establish a one-to-one correspondence between C and R, which implies that if C were a set, it would be bijective with some ordinal A. However, this leads to a contradiction since A cannot be bijective with R. The discussion assumes the framework of Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC). Therefore, the assertion that an unbounded subset of ordinals is a class rather than a set is supported by this reasoning.
RWood
Messages
4
Reaction score
0
Let R be the class of all ordinals. If a subset C of R is unbounded (i.e. for any ordinal \alpha \in R, there is \beta in C with \beta greater than \alpha ), then it seems to me that C cannot be a set, only a class. Is this true, and if so, how does one prove it? My reading on the general subject matter is limited to a bit of web browsing - perhaps the problem is trivial.
 
Physics news on Phys.org
RWood said:
Let R be the class of all ordinals. If a subset C of R is unbounded (i.e. for any ordinal \alpha \in R, there is \beta in C with \beta greater than \alpha ), then it seems to me that C cannot be a set, only a class. Is this true, and if so, how does one prove it? My reading on the general subject matter is limited to a bit of web browsing - perhaps the problem is trivial.

I think I have the outline of a proof (there may of course be something much quicker!).

1) It is quite easy to get a 1-1 correspondence between C and R; a map C=>R is obvious; a 1-1 map R=>C can be constructed by transfinite induction, using
the unboundedness of C to ensure successor elements (or limit ordinals) are mapped to an increasing sequence of C-members.

2) On the other hand, if C is a set then it is bijective with some ordinal A (and some cardinal as well). But then A would be bijective with R, and that is clearly impossible. All this assumes we are a ZFC world.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

Similar threads

Replies
14
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
18
Views
3K
Replies
15
Views
2K
Replies
28
Views
6K
Back
Top