Does a sub vector space based on $\mathbb{I}$ have an infinite basis

Question

I wanted to know whether the following reasoning is correct:

Given a sub vector space $U \subseteq (\mathbb{R}^3, +, \cdot)$ defined as

$$U = \left\{ \begin{pmatrix}a\\b\\c\end{pmatrix} \mid a,b,c \in \mathbb{I} \right\}$$

There exist an infinite amount of elements in $U$ that can't be expressed as a product of another element in $U$ and a scalar in $\mathbb{R}$, i.e. $\forall x \in U, \lambda \in \mathbb{R}. \exists y \in U. \lambda \cdot x \neq y $

Therefore, the basis for the vector space $U$ is $\dim(U)=\infty$

Answer 1

$U$ is not a subvector space of $\mathbb{R}^3$ because it is not closed under scalar mutiplication. For example, let $v=(\pi, \pi,\pi)^T\in U$. Then $\frac{1}{\pi}v=(1,1,1)^T\notin U$, so I think your reasoning doesn't make any sense from the very start.

Answer 2

1. As Javi says, this is not a vector space.

2. For a space to be infinitely dimensional, it's not enough to have an infinite number of vector, each of which is not a scalar product of any other. You have to have an infinite number of vectors, each of which is not a linear combination of all of the other vectors put together. You really need to look at the definition of "dimension".

3. You don't have your qualifiers correct. "$\forall x \in U, \lambda \in \mathbb{R}. \exists y \in U. \lambda \cdot x \neq y $" means that once x and $\lambda$ are fixed, it's possible to find a vector not equal to $\lambda x$, which is a rather trivial claim. You probably meant "$\forall x \in U \exists y \in U \forall \lambda: \lambda \cdot x \neq y $".