If $F$ is a vector subspace of $E$ such that the interior is non empty, then $E=F$.

Question

Let $(E,\|\cdot \|)$ a normed space and let $F$ a subspace of $E$. Prove that if the interior of $F$ is non empty, then $E=F$.

Attempts

Suppose $E$ has finite dimension. Suppose $\dim F<\dim E$. Let $E=Span\{v_1,...,v_n\}$ and suppose WLOG that $F=Span\{v_1,...,v_m\}$ with $m<n$. In particular, if $x=(x_1,...,x_m,0,...,0)\in F$, then for all $r$, $(x_1,...,x_m,\frac{r}{2},0,...,0)\notin F$ will always be in $B(x,r)$, and thus, there is no ball that is included in $F$. Therefore $Int(F)=\emptyset$.

Question : How can I do if $E$ has finite dimension ?

Answer 1

Let $x$ be an interior point of $F$. Then there exists $r>0$ such that $B(x,r) \subset F$. In particular $x \in F$. Hence $||y|| <r$ implies $y+x \in B(x,r) \subset F$ so $y =(y+x)-x \in F$ because $F$ is a subspace. Now let $z$ be any non-zero element of $E$. Take $y=\frac r {2||z||} z$. Then $||y|| =\frac r 2<r$ so $y \in F$. Hence $z=\frac {2||z||} r y \in F$. Since $F$ is a subspace this implies $z \in F$. Thus $E=F$.

Answer 2

An other way would be as following : let $a\in Int(F)$ and $x\in E$, $x\neq a$. Let $r>0$ s.t. for all $z\in E$, if $\|z-a\|<r$, then $z\in F$. Let $y$ s.t. $$y-a=\frac{r}{2\|x-a\|}(x-a).$$ In particular, $y\in F$. Moreover, $$x=a+\frac{2\|x-a\|}{r}(y-a),$$ and thus $x\in F$ as well. The claim follow.