Post History
#2: Post edited
by
Prime Mover
·
2022-06-10T09:57:37Z (over 2 years ago)
- Here's a good one: Every set is well-orderable iff the Axiom of Choice holds.
It is quite challenging to prove that the Axiom of Choice implies that every set is well-orderable. Smullyan and Fitting build up to it in a focused by leisurely manner over the course of 3 chapters in their "Set Theory and the Continuum Problem" (2010, rev. ed.) During the course of this, they establish a number of theorems about classes inductive sets, superinductive sets, progressing functions, slowly progressing functions, constructs they call $g$-towers and slow $g$-towers, and culminate with a straightforward proof which defines a slowly progressing function $g$ based on applying a choice function to the complement of the sequence of images of that function.- However, if every set is well-orderable, the axiom of choice follows trivially. Indeed, let $S$ have a well-ordering. Then for every non-empty subset of a set $S$, the choice function on that subset is its smallest element under that well-ordering.
- Here's a good one: Every set is well-orderable iff the Axiom of Choice holds.
- It is quite challenging to prove that the Axiom of Choice implies that every set is well-orderable. Smullyan and Fitting build up to it in a focused but leisurely manner over the course of 3 chapters in their "Set Theory and the Continuum Problem" (2010, rev. ed.) During the course of this, they establish a number of theorems about classes inductive sets, superinductive sets, progressing functions, slowly progressing functions, constructs they call $g$-towers and slow $g$-towers, and culminate with a straightforward proof which defines a slowly progressing function $g$ based on applying a choice function to the complement of the sequence of images of that function.
- However, if every set is well-orderable, the axiom of choice follows trivially. Indeed, let $S$ have a well-ordering. Then for every non-empty subset of a set $S$, the choice function on that subset is its smallest element under that well-ordering.
#1: Initial revision
by
Prime Mover
·
2022-06-10T09:57:16Z (over 2 years ago)
Here's a good one: Every set is well-orderable iff the Axiom of Choice holds. It is quite challenging to prove that the Axiom of Choice implies that every set is well-orderable. Smullyan and Fitting build up to it in a focused by leisurely manner over the course of 3 chapters in their "Set Theory and the Continuum Problem" (2010, rev. ed.) During the course of this, they establish a number of theorems about classes inductive sets, superinductive sets, progressing functions, slowly progressing functions, constructs they call $g$-towers and slow $g$-towers, and culminate with a straightforward proof which defines a slowly progressing function $g$ based on applying a choice function to the complement of the sequence of images of that function. However, if every set is well-orderable, the axiom of choice follows trivially. Indeed, let $S$ have a well-ordering. Then for every non-empty subset of a set $S$, the choice function on that subset is its smallest element under that well-ordering.