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I imagine no since the dimensions do not match

but they have the same cardinality $|\mathbb R |= |\mathbb C|$?

Theo Bendit
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user29418
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    If the cardinalities are the same then there must be a bijection by definition. – copper.hat Mar 23 '19 at 03:32
  • Maybe something with bit shuffling - choose every other bit for $R \to C$, and interlace the bits for $C \to R$. Making it 1-to-1 seems challenging. – marty cohen Mar 23 '19 at 03:33
  • Did you do any research before asking this question? For example, the very first google hit for the query "bijection r to c" is this MSE post. (-1) –  Mar 23 '19 at 03:41
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    I did do research but the results didn't match. I wanted R->C (which is isomorphic to R^2). – user29418 Mar 23 '19 at 03:42
  • How you are claiming $|\mathbb{R}|=|\mathbb{C}|$? The cardinality is defined by bijection between two sets. – tarit goswami Mar 23 '19 at 03:44

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If you just want a bijection as sets, the answer is yes. That’s exactly what it means for two sets to have the same cardinality. If you’re looking for a bijection that preserves some structure, that will depend on precisely what you’re trying to preserve but the answer is very probably no.

Robert Shore
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