Subspace of a normed vector space equal to It?

Question

If $E$ is a normed vector space and $F$ a subspace of $E$. I'm trying to prove that if F contains $B(x,r)$ then it also contains $B(0,r)$ and $F=E$.

I started by writing the definition: $B(x,r)=\{y: \Vert y-x \Vert<r\}=\{x+v: \Vert v\Vert<r\}$ So $F $contains $x \; \text{and} \; x+v$ then it contains their difference which is $v$ so $F$ contains $B(0,r)$.

Can i just say that F being a subspace contains $0_E$ and $B(0,r)= \{v: \Vert v \Vert < r\}$ which is just a subspace of $B(x,r)$ ?

But i'm stuck with the second part, proving F=E, i don't know where to begin.

Answer 1

The fact that $F$ is a subspace implies two things:

1) It is closed under addition (and contains zero).

2) It is closed under multiplication.

Now, suppose that $B(x,r) \subset F$. This means that $x \in F$, and by closure under scalar multiplication, $(-1)x = -x \in F$.

Now, let $v \in B(0,r)$, then $v +x \subset B(x,r)$, right? So we can say that $v +x =y$ for some $y \in B(x,r)$. But the, $v=y-x$, and $F$ is closed under addition, therefore $v \in F$, because $y \in B(x;r) \subset F$ and $-x \in F$.

This shows that because we did not assume anything about $v$ other than $v \in B(0,r)$, it follows that $B(0,r) \subset B(x,r)$.

To show that $E=F$, let $w \in E$ be a vector. If $w=0$, then it is already in $F$, so we will assume it is non-zero, hence has norm non-zero as well.Then, let $c = \frac{2||w||}r$ (so $c \neq 0$). Note that the vector $\frac 1c w$, has norm exactly equal to $\frac r2$, hence is in $B(0,r)$, hence in $F$. But, by closure under scalar multiplication, $(c)(\frac 1c w)$ is also in $F$, but this is equal to $w$, so $w \in F$. Since we did not assume anything about $w$ other than the fact that it is in $E$, it follows that $E \subset F$. But then $E$ already contains $F$, so $E=F$.

Your second point is wrong: in fact, $B(0;r)$ and $B(x ; r)$ need not even intersect! Furthermore, none of them are vector spaces, they aren't closed under scalar multiplication or addition, and $B(x;r)$ may not even contain zero. Hence, this approach is used in the problem.