Skip to main content
MathOverflow

Return to Answer

added 10 characters in body
Source Link
Stefan Geschke
Stefan Geschke
  • 16.2k
  • 2
  • 54
  • 82

Short answer: No.

By countably infinite subset you mean, I guess, that there is a 1-1 map from the natural numbers into the set.

If ZF is consistent, then it is consistent to have an amorphous set, i.e., aan infinite set whose subsets are all finite or have a finite complement. If you have an embedding of the natural numbers into a set, the image of the even numbers is infinite and has an infinite complement. So the set cannot be amorphous.

Short answer: No.

By countably infinite subset you mean, I guess, that there is a 1-1 map from the natural numbers into the set.

If ZF is consistent, then it is consistent to have an amorphous set, i.e., a set whose subsets are all finite or have a finite complement. If you have an embedding of the natural numbers into a set, the image of the even numbers is infinite and has an infinite complement. So the set cannot be amorphous.

Short answer: No.

By countably infinite subset you mean, I guess, that there is a 1-1 map from the natural numbers into the set.

If ZF is consistent, then it is consistent to have an amorphous set, i.e., an infinite set whose subsets are all finite or have a finite complement. If you have an embedding of the natural numbers into a set, the image of the even numbers is infinite and has an infinite complement. So the set cannot be amorphous.

Source Link
Stefan Geschke
Stefan Geschke
  • 16.2k
  • 2
  • 54
  • 82

Short answer: No.

By countably infinite subset you mean, I guess, that there is a 1-1 map from the natural numbers into the set.

If ZF is consistent, then it is consistent to have an amorphous set, i.e., a set whose subsets are all finite or have a finite complement. If you have an embedding of the natural numbers into a set, the image of the even numbers is infinite and has an infinite complement. So the set cannot be amorphous.