# Cardinality, bijections, infinite sets

We know that if there is a bijection from finite set $$A$$ to finite set $$B$$, we can remove one element from both sets and still be able to form a bijection. Similarly, removing sets of equal cardinalities from finite sets still allows us to form bijections. But this result does not hold for infinite sets.

Let the removed infinite sets be $$A'$$ and $$B'$$, such that they are both proper subsets of $$A$$ and $$B$$ respectively. Let $$A=B=B'=\mathbb{N}$$. Let $$A'$$ be the set of odd numbers. Then, $$A-A'$$ is the set of even numbers. The cardinality of $$(A-A')$$ is $$\mathbb{N}$$. $$B-B'= \emptyset$$ so the cardinality of $$(A-A')$$ is not equal to $$(B-B')$$. Hence, I have proved that the finite set rule I described above does not hold for infinite sets. Is this correct?

• what do you mean let $A=B=B'=N$, if the sets are equal in the first place, then why bother with any of this? Do you mean the cardinalities are equal? Also isn't B' a proper subset of B. – IntegrateThis Oct 4 '18 at 5:13
• The first is statemenr is true for non-empty finite sets. – Thomas Andrews Oct 4 '18 at 5:14
• Because I have set out to prove that for infinite sets, cardinalities of (A-A') and (B-B') might not be equal, I am saying as a counterexample, that A=B'=B=N. – childishsadbino Oct 4 '18 at 5:15
• Seems correct to me. – Ovi Oct 4 '18 at 5:18
• Actually one small mistake. You say that $B'$ is a proper subset of $B$, but then you say $B = B'$. To fix this, you can just let $B' = \mathbb{N} - \{0 \}$. – Ovi Oct 4 '18 at 5:23