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Can you construct a 1-1, onto map from [0,1] to (0,1]? I know there are similar questions being asked but I have not found any help or advice for this specific 1-1, onto map and am rather stuck. Thank you.

Christopher
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  • With Axiom of Choice, easily. – mniip Oct 13 '17 at 05:32
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    Several duplicates at hand... – Did Oct 13 '17 at 05:39
  • @mniip It would be interresting to se how you actually use the AoC for this. It's just not to prove that there exists such a mapping (which we could do with the Schröder-Bernstein theorem). – skyking Oct 13 '17 at 06:20
  • @skyking I had an idea similar to Casper's, but instead of $2^{-n}$ I'd use $1/n$. Suppose $P(x, y) = (\exists n. x=1/n \land y=1/(n+1)) \lor((\not\exists n. x=1/n) \land x=y)$. AoC lets us turn the evident $\forall x. \exists y. P(x, y)$, into an actual function $\exists f. \forall x. P(x, f(x))$. – mniip Oct 17 '17 at 01:49
  • @mniip You definitely don't need AoC for that. You only need ASoS and to prove that $y$ is unique (in order to make the relation a function) which should be rather straight forward. If you resort to AoC you haven't actually constructed such a mapping, but only proved that such a mapping exists. – skyking Oct 17 '17 at 05:18
  • @skyking Indeed you are correct. I seem to have misinterpreted the usage of AoC in another construction similar to this one. – mniip Oct 18 '17 at 07:39

1 Answers1

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Map 0 to $\frac12$, map $\frac12$ to $\frac14$, map $\frac14$ to $\frac18$, etc., and map everything else to itself.

Casper
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