Why is $\dim_{\mathbb{Q}}\mathbb{R}^n=\operatorname{card}\mathbb{R}$ for all $n\geq1$?

Asked Sep 02 '16 at 14:54

Active Sep 02 '16 at 14:54

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Question: How to show that the dimension of $\mathbb{R}^n$, when viewed as a vector space over $\mathbb{Q}$, is equal to the cardinality of $\mathbb{R}$?

This is related to my previous question. This dimension is clearly infinite, but why is it precisely the cardinality of $\mathbb{R}$?

edited Apr 13 '17 at 12:20

Community

asked Sep 02 '16 at 14:54

Paul

For any infinite-dimensional $k$-vector space $V$ and for any $n ≥ 1$, we have $\dim_k(V^n) = \dim_k(V)$. – Watson Sep 02 '16 at 14:57
This is true also for vector spaces with a countable basis. – Bernard Sep 02 '16 at 15:02

Why is $\dim_{\mathbb{Q}}\mathbb{R}^n=\operatorname{card}\mathbb{R}$ for all $n\geq1$?

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