For any given $n$ determine the cardinality of the Euclidean $n$-space $$\mathbb{R}^n = \left\{(x_1, x_2, \dots, x_n) : x_i \in \mathbb{R} \text{ for all } i = 1, 2, \dots, n\right\}.$$ Are there ”more” points in the Euclidean plane $\mathbb{R}^2$, the Euclidean 3-space $\mathbb{R}^3$ or hyperspace $\mathbb{R}^4$ as on the real line $\mathbb{R}$? Can anyone solve it so to understand it? I didn't understand anything.
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For any natural number $n \geqslant 1$, $$|\mathbb{R}^n|=|\mathbb{R}|=\aleph $$ where $\aleph$ denotes the cardinality of the continuum.
EDIT: A cardinality based proof is as follows $$|\mathbb{R}^n|=|\mathbb{R}|^n=(2^{\aleph_0})^n=2^{\aleph_0 \times n}=2^{\aleph_0}=|\mathbb{R}|. $$
user1337
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so the final answer is that all of them have the same cardinality because |Rn|=|R|=ℵ? – soc Dec 07 '17 at 04:16
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@RossMillikan so if we know the R^2-->R we know directly R^3-->R because bijective is one-to-one – soc Dec 07 '17 at 04:31