I have noticed that in order to describe 3-dimensional space you need at least three variables in every Koordinate-System. My question is whether it is possible to assign every single point in n-dimensional space a unique number - in other words whether any bijective function $f:\Bbb{R^n}\rightarrow\Bbb{R} , n\in\Bbb{N}/\{1\}$ exists. If $f$ does exist - what would it look like - and if it doesnt - how would you prove it?
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basically both spaces have the same cardinal – Tsemo Aristide Mar 20 '20 at 14:31
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Remember to use the notation $\backslash$ (in LaTeX \backslash) when you are referring to complement sets. – justadzr Mar 20 '20 at 14:33
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Such a bijection exists. However if you require also that it be continuous, then it does not exist. – GEdgar Mar 20 '20 at 14:34
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See https://en.wikipedia.org/wiki/Cardinality_of_the_continuum – Ethan Bolker Mar 20 '20 at 14:34
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Does this answer your question? Bijection from $\mathbb R$ to $\mathbb {R^N}$ – smalldog Mar 20 '20 at 14:39
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Yes it does! Thank you very much!! – Susp1cious Mar 20 '20 at 14:41
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A bijection $f : \mathbb R^2 \to \mathbb R$ that almost works is obtained by intertwining the decimals. For real numbers $a,b$ written in decimal as \begin{align} a &= \cdots a_{-2} a_{-1} a_0 . a_1 a_2 a_3 \cdots \\ b &= \cdots b_{-2} b_{-1} b_0 . b_1 b_2 b_3 \cdots \end{align} where all but finitely many digits to the left of the decimal are zero, let $$ f(a,b) = \cdots a_{-2} b_{-2}a_{-1}b_{-1} a_0 b_0. a_1 b_1 a_2 b_2 a_3 b_3\cdots $$ I say it almost works because there is a problem involving real numbers with two different decimal expansions, like $1 = 0.999\cdots$. That problem can be fixed, but it is complicated to do it, so I will not do it here. (You can see, instead, the many duplicates of this question.)
As Kenta notes, this also doesn't work because of negative numbers.
GEdgar
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