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Is there any bijection between $\mathbb{R}$ and $\mathbb{R}^2$ ? If have then what is the mapping ? Please define the mapping. They have same cardinality then it is possible to have a bijection between them.

qualcuno
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sudip singha
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1 Answers1

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Hint: first, show that $\mathbb{R} \simeq (0,1)$. Then, observe that if $(0,1) \simeq (0,1)^2$, then $\mathbb{R} \simeq \mathbb{R}^2$. For the former, one can consider a map such as $0.a_1 a_2 a_3 a_4 \dots \mapsto (0.a_1 a_3 \dots,0.a_2a_4\dots) $.

qualcuno
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    This needs a little cleanup to deal with numbers that have two decimal expansions. You can't just ignore the ones that end in all $9$s, for example, because you have numbers in $\Bbb R$ like $0.1929394959\ldots$ that create that kind of number. – Ross Millikan Mar 20 '19 at 04:55
  • @RossMillikan indeed so. I think that picking a convention fixes it though, as for example taking $0.a_1 \dots a_N 9999 \dots$ with $a_N < 9$ to be always represented as $0.a_1 \dots (a_N +1) 000 \dots$. – qualcuno Mar 20 '19 at 04:57
  • No, that is not enough. Consider the two numbers in $\Bbb R$, $0.11000000\ldots$ and $0.1090909090909\ldots$. They do not have other forms, but they map to $(0.1,0.1)$ and $(0.1,0.09999\ldots)$, which are the same point. If you choose one form in $\Bbb R^2$ you miss some points of $\Bbb R$. The best I have found is to do the interleaving on all digit strings, then to find a bijection between the digit strings and $\Bbb R$. Now you only have countably many problems to deal with and you can use the "send them down the line" trick in various ways. – Ross Millikan Mar 20 '19 at 14:20
  • @RossMillikan I see, that seems to do the trick. Thanks! – qualcuno Mar 20 '19 at 16:51
  • You can also just do an injection in each direction and apply to Schroeder-Cantor-Bernstein. For $\Bbb R^2 \to \Bbb R$ you can then just choose one representation. You will miss some points in $\Bbb R$ as above, but that is no longer a problem. Many people don't see that as "defining a mapping", but the theorem is constructive. It is just hard to say where some points go. – Ross Millikan Mar 20 '19 at 16:56
  • @RossMillikan yeah, I guess that works too, but what's a constructive proof of the theorem? The only one I know uses a fixed point theorem on complete lattices and (iirc) is not very constructive. – qualcuno Mar 20 '19 at 17:29
  • The one I am familiar with is similar to the one in Wikipedia. It explicitly creates a bijection by saying which of the injections to use for each element. – Ross Millikan Mar 20 '19 at 17:35
  • @RossMillikan Interesting! I will check it out. – qualcuno Mar 20 '19 at 17:45