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!

  • 6
    $\begingroup$ Relax. It's fine. $\endgroup$ Sep 22, 2018 at 10:12
  • $\begingroup$ Thank you @JoséCarlosSantos :) I'm really relaxed ^^ $\endgroup$
    – Akira
    Sep 22, 2018 at 12:43


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