Why does |A| = |B| IFF A bij B

Question

I don't understand why the mapping lemma states that |A| = |B| IFF A bij B.

My understanding of the above lemma is

If A bij B is True then |A| = |B| is True

If A bij B is False then |A| = |B| is False

If |A| = |B| is True then A bij B is True

If |A| = |B| is False then A bij B is False

My understanding is most likely incorrect so please correct my understanding of the mapping lemma:

This mapping lemma is invalid by providing this counter example. A = {1,2,3}. B = {4,5,6}. There is no relation between set A and B.

With this example, |A| = |B| is True since 3 = 3. However A bij B is False since there is no relation between them. This means that in this counter example, (|A| = |B| IFF A bij B) = (T IFF F) = F

I am almost certain my understanding and counter example are wrong but can someone kindly point out where I made a mistake in my understanding and example? Thank you.

Answer 1

Based on your comment under Kenny Lau's answer, I think the problem is the following: that you are assuming that a bijection between two sets needs to be "meaningful" in some way. This is not the case: bijections don't need to follow some pattern, they're allowed to be completely arbitrary.

Indeed, the existence of a bijection between two sets is usually how "having the same cardinality" is defined; really what we're doing here is arguing that this formal definition matches our usual intuitions about cardinality.

There is a philosophical point here - to what extent can mathematical objects "exist" without being "definable" - but it's not really relevant here: two sets have the same cardinality iff there is a bijection between them, regardless of what "is" means. Perhaps two different notions of mathematical existence could disagree about whether a bijection exists between two particular sets, but both notions will separately agree that "equicardinality = bijection."

Answer 2

$x \mapsto x + 3$ is a bijection $A \to B$.

Answer 3

Another perspective on a bijection is that it is a function F: X -> Y, such that there exists an inverse function G: Y -> X. So, for every element x in X, we have G(F(x)) = x, and for every element y in Y we have F(G(y)) = y.

Cardinality is trichotomous in that either |X| < |Y|, |X| = |Y|, or |X| > |Y|.

So, for a set X with a bijection F from X to Y, suppose that |X| < |Y|. But, then there exists some y in Y such that F(G(y)) = y is false. Thus, F is not a bijection. We have a contradiction, so |X| = |Y| or |X| > |Y|.

Suppose that for the same set with a bijection F from X to Y that |X| > |Y|. But, then there exists some x in X such that G(F(x)) = x is false. Thus, F is not a bijection. We have a contradiction, and we conclude that |X| = |Y|.