Is R×R⊆R?
If this is the case then would it be true that |R×R|≤|R|?
Answer
R×R⊆R is incorrect.
It is incorrect for the same reason that in vector spaces R3⊆R2 is incorrect. The number of components is different.
However...
The statement |R×R|≤|R| is correct. and infact, it is true that |R×R|=|R|
A simple proof for |R×R|≤|R| without resorting to cardinal arithmetique would be to find a function f:R2→R that is injective. Can you think of such function? how about f(i,j)=2i3j?
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