For infinite sets and to be numerically equivalent. Is bijective function
the only requirement that needs satisfying?
There is apparently another way of comparing set cardinalities, with cardinal numbers but I never learned anything on the topic and someone more knowledgeable on the matter will have to discuss it.
(Note that when I say if and only if its merely a logical statement used in every definition, sometimes implicitly.)
In my question, and are infinite.
I know this theorem: An infinite subset of a denumerable set is denumerable.
But I am tweaking it for a different scenario, where I know that neither nor is denumerable, except there is a bijective function from A to B.
To make it simple. Suppose that and
is defined by
Since we know that is bijective, despite the fact that and being uncountable, is it sufficient to conclude that ?
Drexel runs at hypersonic speed when I am still crawling. I only knew that if is injective, the , but I couldn't see how that automatically translate to being bijective.
Now, I think you mean if is bijective, then , and if is surjective, then . Then only after putting those two together,
For example, there clearly exists an injection , this is just . So to prove that you would just need to find an injection from to , you don't need to go the whole hog and find a bijection!
Moreover, if A = the empty set, and B = the empty set, then you don't need set theory to prove |A| = |B|, since it follows by identity theory alone (A = the empty set = B, so A = B, so |A| = |B|).
By the way, I think in some of the discussion above it was taken as implicit that if there is a surjection from A onto B then there is an injection from B into A. However, just to note, that requires the axiom of choice.
The salient principles here that do not require the axiom of choice:
By definition, A and B are equinumerous if and only if there is a bijection from A onto B.
And, if there is a bijection from A onto B, then there is a bijection from B onto A (just take the inverse of the bijection from A onto B).
And, if there is an injection from A into B and there is an injection from B into A, then there is a bijection from A onto B. And, such a bijection can be recovered from the two given injections (by following any of the constructive proofs of Schroder-Bernstein).
Theorem: defines a linear ordering on the set of all cardinal numbers.
Proof: The only thing that is unclear is the linearity of . So, we must prove that given sets it is true that . If either or is empty the result is clear, so assume not. Fix and and define in the only possible way, clearly this is an injection. Let . Define the ordering on by . This clearly is a partial ordering. We now must prove that every chain has an upper bound. But, it is clear (I hope) that where is defined in the obvious way (@novice there's an exercise, figure it out how to say that in a rigorous manner). Also, once you have defined it will be clear that is an upper bound for from where it follows from Zorn's Lemma that must have a maximal element . Now, suppose that . Then, choosing we may define by and and then of course would contradict the maximality of .
Thus, there exists a bijection from one of or into a subset of the other, and the conclusion follows.
I don't think I said that very eloquently, but oh well.
In essence all your doing is pairing off elements until you run out.