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Prove that if $n \in \Bbb N$, $f: I_n \to B$, and $f$ is onto, then $B$ is finite and $|B|\leq n$.

Notes on notation:

For each natural number $n$, $I_n = \{i \in \mathbb{Z} \mid i \leq n\}$.

$A \sim B$ indicates that $A$ is equinumerous to $B$.

$f: I_n \rightarrow B$ means there is a function from $I_n$ to $B$.

  • 0
    I think you mean that $I_n=\{i\in\Bbb Z^+:i\le n\}$.2012-12-03
  • 2
    What have you got so far? Which bit did you get stuck on?2012-12-04

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