Prove if $A$ is infinite and $B$ is finite, then card($A \cup B$) = card($A$).

Question

Can anyone provide me with function and detail instruction Please.

Answer 1

Hint: Any infinite set has a subset that is countable. Show that your statement is true when $A$ is countable, then show it for general infinite $A$.

Answer 2

Hint: Show that for every finite $n$, $\Bbb N$ has the same cardinality as $\Bbb N\cup\{-1,\ldots,-n\}$. Conclude that you can use $B$ instead of $\{-1,\ldots,-n\}$. Now use the fact that $A$ is infinite and therefore has a subset $A_0$ such that $A_0$ is equipotent with $\Bbb N$, and conclude the same is true for $A$ and a finite set $B$.

Alternatively: Prove the above for the case when $B$ is a singleton, and use induction to conclude that the result is true for any finite number of elements.

Answer 3

One can avoid axiom of (countable) choice as follows:

If $A$ is infinite, then by definition $A$ is equipotent with a proper subset $A_0$ of $A$. Now we can use Cantor-Bernstein-Schroeder theorem to see that $A$ and $A_0 \cup \{ p \}$ are equipotent where $p \in A - A_0$. Then, we can proceed using induction.