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?

  • $\begingroup$ 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. $\endgroup$ – IntegrateThis Oct 4 '18 at 5:13
  • $\begingroup$ The first is statemenr is true for non-empty finite sets. $\endgroup$ – Thomas Andrews Oct 4 '18 at 5:14
  • $\begingroup$ 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. $\endgroup$ – childishsadbino Oct 4 '18 at 5:15
  • $\begingroup$ Seems correct to me. $\endgroup$ – Ovi Oct 4 '18 at 5:18
  • $\begingroup$ 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 $B' = \mathbb{N} - \{0 \}$. $\endgroup$ – Ovi Oct 4 '18 at 5:23

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