The following statements are well known:
- A space is finite dimensional if and only the units sphere is compact.
- The dimension of the normed space $\mathbb{R}^n$ over the field $\mathbb{Q}$ is $\infty$.
Does this mean that the unit sphere in $\mathbb{R}^n$ over $\mathbb{Q}$ compact?
The space $\mathbb{R}^n$ examined over the field $\mathbb{Q}$ is just like $\mathbb{R}^n$ (all vectors with $n$ real components), but the scalars for scalar multiplication, i.e. in linear combinations can only be rational. The unit sphere are all vectors whose norm is less than or equal to one.
I thought that this could be true since the unit ball is only dependent on the norm we use and even though we examine $\mathbb{R}^n$ over a difference filed than usual, we can choose the norm which is induced by the scalar product, i.e. $\| x \| := \sqrt{\sum_{k = 1}^{n} x_k^2}$, yielding the same unit ball as if we would examine $\mathbb{R}^n$ over $\mathbb{R}$.
Is this reasoning correct? Have I missed any details in the above statements, which make my conclusions false?
I also know that for every vector space we can create a norm by summing the absolute values of the coefficients of a vector when writing it out as linear combination of its base elements, but I don't have to choose that norm, right?
In finite dimensions, all norms are equivalent, so I can just choose one. But what is the "natural norm" on such an weird space?
Best Answer
My confusion arose from the fact that statement 1 only holds for normed spaces, which by definition have $\mathbb{R}$ or $\mathbb{C}$ as their base fields. This is not the case of my space, therefore the result is not applicable.