A function $f$ having , for all $k \in N$, the subsets of $A$ given by the solutions to $f(a)=k$ is finite, show that $A$ is countable or finite.

Question

Could not solve this question, could anyone discuss it with me please?

Answer 1

Let $A_k = \{a\in A : f(a)=k\}$ so by assumption $A_k$ is finite. Now $A=\bigcup_{k=1}^\infty A_k$, it follows that $A$ is a countable union of finite sets and therefore is finite or countable.

Answer 2

More detail. Let $a \in A$. Let $f(a) = k \in \mathbb N$. Then $a \in \{b \in A| f(b) = k\}$. So $a \in \cup_{k\in \mathbb N}\{b \in A| f(b) = k\}$. So $A \subset \cup_{k\in \mathbb N}\{b \in A| f(b) = k\}$. And if $c \in \cup_{k\in \mathbb N}\{b \in A| f(b) = k\}$ then $c \in \{b \in A| f(b) = k\}$ for some $k$ and so $c \in A$. So $\cup_{k\in \mathbb N}\{b \in A| f(b) = k\}\subset A$.

So $A = \cup_{k\in \mathbb N}\{b \in A| f(b) = k\}$.

So $A$ is a subset of a countable union of finite sets.

You should have theorem that the countable union of finite sets is at most countable.