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How do we define this mapping:

Let $A$ be a infinite set and let $B \subset A$. Suppose that $B$ is finite. Show that there exists a bijective mapping $ $ $f: A \rightarrow A\setminus B$

We get a hint: Use denumeration $B = \left\{ b_1, b_2, ...,b_n \right\}$

Helgi Freyr Ásgeirsson
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  • Is the set countably infinite or just not finite? – JMJ May 02 '17 at 23:55
  • I assume that it is not finite according to our defenitions – Helgi Freyr Ásgeirsson May 03 '17 at 00:11
  • Possibly related: https://math.stackexchange.com/questions/25241/proof-there-is-a-1-1-correspondence-between-an-uncountable-set-and-itself-minus. I will be the first to admit I didn't believe the assertion at first for the uncountable case but failed to find a counterexample. Evidently it is true. – JMJ May 03 '17 at 00:44
  • This cannot be done without the Axiom of Choice. – DanielWainfleet May 03 '17 at 01:58

1 Answers1

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If $A$ is countably infinite, write $$ A = \{a_{1} = b_{1}, a_{2} = b_{2}, \ldots, a_{n} = b_{n}, a_{n+1}, a_{n+2}, \ldots \} = B \cup \{ a_{n+1}, a_{n+2}, \ldots \} $$ and define $f(a_{k}) = a_{k + n}$.

avs
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