For infinite sets and to be numerically equivalent. Is bijective function
the only requirement that needs satisfying?
I'm not sure what you mean by numerically equivalent. One way of defining two sets to have the same cardinality if and only if there exists a bijection between the two.
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.)
Are you thinking of and being countable?
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
or equivalently
is defined by
Since we know that is bijective, despite the fact that and being uncountable, is it sufficient to conclude that ?
Well, you have made it very clear for me to understand.
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,
and .
Yah?
Yes, if surjective then this implies . However, the general technique is that if is an injection and is also an injection, then and so .
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!
There is a bijection from the empty set onto the empty set. The empty set itself is a bijection from the empty set onto the empty set.
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).
I would like to remark that just like the above remark there is another thing which needs to be proven using Zorn's Lemma (AKA AOC).
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.