Interior of Proper Subspace of Normed Vector Space is Empty

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.

Let $U$ be a proper vector subspace of $X$.

Let $U^\circ$ be the interior of $U$.

Then $U^\circ = \O$.

Proof
Suppose that $U^\circ \ne \O$.

Take $x \in U^\circ$.

Then there exists $\epsilon > 0$ such that $\map {B_\epsilon} x \subseteq U$, where $\map {B_\epsilon} x$ is the open ball centered at $x$ with radius $\epsilon$.

Since $U$ is a vector subspace, we have $\map {B_\epsilon} 0 = \map {B_\epsilon} x - x \subseteq U$.

So if $x \in X$ has $\norm x < \epsilon$, then $x \in U$.

Take $x \in U \setminus \set {\mathbf 0_X}$.

Then, we have:


 * $\ds \norm {\frac \epsilon {2 \norm x} x} = \frac \epsilon 2 < \epsilon$

so that:


 * $\ds \frac \epsilon {2 \norm x} x \in U$

Since $U$ is a vector subspace, we have $x \in U$.

We also have $\mathbf 0_X \in U$, so $U = X$.

This is contrary to the hypothesis that $U \ne X$.

So we have $U^\circ = \O$.