How would we prove that infinite sets have at least a cardinality of aleph naught?
All infinite sets have a cardinality of at least aleph naught
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0@Martin: Yes. I suppose. I am also certain that this question has been answered uncountably many times on this site before. – 2012-09-30
3 Answers
I am not sure to which extent this fulfills OPs needs, but perhaps it is useful to have this somewhere for reference. (I guess this can be found, in some form, in several other questions at this site.)
Several definitions of finite and infinite sets are used in mathematics. The following result, taken from H. Herrlich: Axiom of Choice, p.44, shows that one of them, called Dedekind-infinite, is equivalent to having cardinality at least $\aleph_0$. You can find a detailed proof there.
Definition 4.1. A set X is called Dedekind–infinite or just D–infinite provided that there exists a proper subset $Y$ of $X$ with $|X| = |Y|$; otherwise $X$ is called Dedekind–finite or just D–finite.
Proposition 4.2. Equivalent are:
(1) $X$ is D-infinite;
(2) $|X|=|X|+1$;
(3) $\aleph_0 \le |X|$.
Every infinite set A has a sequence {a1,a2,a3,...} in it because if it doesn't, it implies that the number of elements in the set is not infinite. Now there is a bijection between the sequence and the set of Natural numbers i.e. a1 <--> 1, a2 <--> 2, a3 <--> 3 and so on. Now A{a1,a2,a3,...} might be empty or not.
If it is empty it implies that cardinality(A) = cardinality(N) where cardinality(A) means the number of elements in the set A and N is the set of Natural numbers.
If it is not empty then definitely cardinality(A) > cardinality(N). Hence cardinality(A) >= cardinality(N) (or aleph naught).
Let A be a set such that it has a cardinality of aleph naught (|A| = ℵ0). Let B be a set and B is a subset of A. Since A is countable we can list its elements as A = {a1, a2, a3,...}. Now we may list the elements of B beginning with the smallest which is in B: a1, a2, a3, ... We either run out of elements in which case B is finite, or we don't (that means that |B| = ℵ0). Thus, ℵ0 is the smallest infinite cardinality because the biggest cardinality below it can only be finite.