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It is easy to show that there is bijection from $\mathbb N$ to $\mathbb N^2$, but is there similar one for $\mathbb R$ and $\mathbb R^2$ ?
This problem is equivalent to other one - "Is there function $f:\mathbb R^2 \rightarrow \mathbb R$ that has inverse of the form $f^{-1}:\mathbb R \rightarrow \mathbb R^2$?"
Thanks for all the help

Jakub Pawlak
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    What does it mean "countable over"? – Asaf Karagila May 21 '20 at 20:53
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    What does "countable over $\mathbb R$" mean? If you are thinking about vector spaces, then $\mathbb R^2$ has dimension $2$ over $\mathbb R$. – lulu May 21 '20 at 20:53
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    If you mean, “Is $\mathbb{R}^2$ bijectable with $\mathbb{R}$?” the answer is “yes”. – Arturo Magidin May 21 '20 at 20:53
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    Note that while $\mathbb N$ is countable, and has the same cardinality as $\mathbb N \times \mathbb N$, $\mathbb R$ is not countable, but has the same cardinality as $\mathbb R \times \mathbb R$, hence, $\exists$ a bijection between $\Bbb R$ and $\Bbb R \times \mathbb R$.. Please see Arturo's comment immediately above. – amWhy May 21 '20 at 20:55
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    regarding a bijective map from $\mathbb R^2$ to $\mathbb R$, see this question – J. W. Tanner May 21 '20 at 20:56

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