Equivalent Conditions for Dedekind-Infinite Set
Theorem
For a set $S$, the following conditions are equivalent:
- $(1): \quad$ $S$ is Dedekind-infinite.
- $(2): \quad$ $S$ has a countably infinite subset.
The above equivalence can be proven in Zermelo-Fraenkel set theory.
If the axiom of countable choice is accepted, then it can be proven that the following condition is also equivalent to the above two:
- $(3): \quad$ $S$ is infinite.
Proof
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(1) implies (2)
Let S be a Dedekind-infinite set.
This means that S is equilvant to a proper subset of itself.
That means there exists a map F from S to a proper subset of itself.
Let A be an element of S not in the image of F.
Also let A_n be a sequence such that A_0 = A and A_(n+1) = F(A_n).
The elements of A_n form a countably infinite set.
(2) implies (1)
Let S be a set with a countably infinite subset A.
We can define F such that if x is an element of A, then f(x) = f(A_n) = f(A_(n+1)) (and f(x) = x otherwise)
Thus F is a bijection between S and a subset of S (S minus A_0).
This means it is equilvant to that subset.
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