Let $E$ a normed vector space and $L \neq E$ a vector subspace of $E$. Can $L$ contain any open ball of $E$?

Question

I was trying to solve a problem of the book Elementos de Topologia Geral by Elon Lages and I found an exercise that I don't have success. The exercise is: prove that any vector subspace $L≠E$ of a normed vector space $E$ couldn't contain any open ball of $E$. Could someone give me a hint? Thank you.

Answer 1

Think about a ball for a moment. There is a point in the centre, and then there are vectors (possibly short ones) from that point in all directions. If a ball is contained in a subspace, then all of those directions are contained in the subspace. Since the subspace is closed under scalar multiplication, we can stretch these direction vectors as much as we like, until we get every point in the space.

So, let's prove this. Let's suppose that $L$ is a subspace of $E$, and that $B[y; r] \subseteq L$. I aim to show that $L = E$. We already know $L \subseteq E$, so let's show $E \subseteq L$.

Suppose $x \in E$. I want to shrink down the vector from $0$ to $x$ until it is length $r$. The resulting vector is $r\frac{x}{\|x\|}$. Consequently, $y + r\frac{x}{\|x\|} \in B[y; r] \subseteq L$. Further, $y \in B[y; r] \subseteq L$, so $$r\frac{x}{\|x\|} = \left(y + r\frac{x}{\|x\|}\right) - y \in L,$$ since $L$ is closed under subtraction. Since it is also closed under scalar multiplication, $$x = \frac{\|x\|}{r}\left(r\frac{x}{\|x\|}\right) \in L$$ as well. Therefore, $E = L$ as required.