Correct option is C
Solution:
Option (a):
"S is in bijection with the set of rational numbers."
- The set of rational numbers (Q) is countable. If S is countable, S could be in bijection with Q. However, the question does not specify that S is countable; S could also be uncountable.
- Therefore, this is not always true.
Option (b):
"S is in bijection with the set of real numbers."
- The set of real numbers (R) is uncountable. If S is uncountable and of the same cardinality as R, then S could be in bijection with R. However, if S is countable, it cannot be in bijection with R.
- Therefore, this is not always true.
Option (c):
"S is in bijection with S × S."
- The axiom of choice ensures that any infinite set S is in bijection with its Cartesian product S × S. This is because the cardinality of S remains unchanged when paired with itself, as shown by set theory.
- Therefore, this statement is always true.
Option (d):
"S is in bijection with the power set of S."
- The power set of S (P(S)) always has a strictly greater cardinality than S itself (by Cantor’s theorem). Hence, S cannot be in bijection with P(S).
- Therefore, this is false.
Conclusion: The correct answer is (c): S is in bijection with S × S.
