Post History
#2: Post edited
by
Michael Hardy
·
2024-07-17T14:51:16Z (3 months ago)
The nonstandard models would be elementarily equivalent to the standard models, i.e. they would satisfy the same first-order formulas. In particular, for every first-order formula, the set of numbers satisfying it would have a least element.- However, for every number $x>0,$ the number $x-1$ exists. Thus for every nonstandard number $x$, the number $x-1$ exists. It is a nonstandard number. Hence there can be no minimal nonstandard number.
- Hence we must conclude that the set of all nonstandard numbers cannot be defined by any first-order formula.
- The nonstandard models would be elementarily equivalent to the standard models, i.e. they would satisfy the same first-order formulas. In particular, for every first-order formula with one free variable, the set of numbers satisfying it would have a least element.
- However, for every number $x>0,$ the number $x-1$ exists. Thus for every nonstandard number $x$, the number $x-1$ exists. It is a nonstandard number. Hence there can be no minimal nonstandard number.
- Hence we must conclude that the set of all nonstandard numbers cannot be defined by any first-order formula.
#1: Initial revision
by
Michael Hardy
·
2024-07-17T04:44:31Z (3 months ago)
The nonstandard models would be elementarily equivalent to the standard models, i.e. they would satisfy the same first-order formulas. In particular, for every first-order formula, the set of numbers satisfying it would have a least element. However, for every number $x>0,$ the number $x-1$ exists. Thus for every nonstandard number $x$, the number $x-1$ exists. It is a nonstandard number. Hence there can be no minimal nonstandard number. Hence we must conclude that the set of all nonstandard numbers cannot be defined by any first-order formula.