Post History
#3: Post edited
by
r~~
·
2022-09-05T05:34:32Z (over 1 year ago)
Add reference
- The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
- In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
- $$
- \|\mathbf{V}\| = \begin{cases}
- \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
- \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
- \end{cases}$$
In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
- The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
- In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
- $$
- \|\mathbf{V}\| = \begin{cases}
- \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
- \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
- \end{cases}$$
- In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
- (For a quickie proof of the infinite-dimensional case, see <https://math.stackexchange.com/a/194287>.)
#2: Post edited
by
Derek Elkins
·
2022-09-05T05:04:16Z (over 1 year ago)
cases environment accomplishes the same output in a simpler way
- The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
- In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
- $$
\|\mathbf{V}\| = \left\\{\begin{array}{ll}- \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
- \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
\end{array}\right.$$- In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
- The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
- In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
- $$
- \|\mathbf{V}\| = \begin{cases}
- \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
- \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
- \end{cases}$$
- In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
#1: Initial revision
by
r~~
·
2022-09-04T03:41:39Z (over 1 year ago)
The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
$$
\|\mathbf{V}\| = \left\\{
\begin{array}{ll}
\|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
\max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
\end{array}
\right.
$$
In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.