# Suppose that $f:A\to B$ where $A$ is uncountable and $B$ is countable. Then there exists $b \in B$ such that $f^{−1}[\{b\}]$ is uncountable.

Suppose that $$f$$ is a mapping from an uncountable set $$A$$ to a countable set $$B$$. Then there exists $$b \in B$$ such that $$f^{−1}[\{b\}]$$ is uncountable.

My attempt:

Lemma: Countable union of countable sets is countable. (I presented a proof here)

Assume the contrary that $$f^{−1}[\{b\}]$$ is countable for all $$b \in B$$. It's clear that $$A=\bigcup\limits_{b\in B}f^{−1}[\{b\}]$$. Moreover, $$\bigcup\limits_{b\in B}f^{−1}[\{b\}]$$ is a countable union of countable sets. Then $$A=\bigcup\limits_{b\in B}f^{−1}[\{b\}]$$ is countable by Lemma. This clearly contradicts the fact that $$A$$ is uncountable. Hence there exists $$b \in B$$ such that $$f^{−1}[\{b\}]$$ is uncountable.

Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

• Relax. It's fine. Sep 22, 2018 at 10:12
• Thank you @JoséCarlosSantos :) I'm really relaxed ^^ Sep 22, 2018 at 12:43