# Interior of linear subspace

Suppose that $$H$$ is a Hilbert space and $$K$$ is a closed linear (strict) subspace of $$H$$. Then, is $$\operatorname{int}(K)=\emptyset?$$

This seems to be the case, for example when I take $$K\triangleq \mathbb{R}^1$$ and $$H\triangleq \mathbb{R}^2$$, then there is no open ball contained in $$K$$; since the open balls generate the Euclidean topology, then the interesection of all open subsets contained in $$K$$ must only be the empty set....Am I wrong?

If you have $$B(x, \epsilon)$$ in your subspace, then you have $$x$$, so you have $$B(0, \epsilon) = B(x, \epsilon) - x$$
And $$B(0, \epsilon)$$ contains a multiple of every vector in the space, so it generates the full space.
If $$x$$ is an interior point of $$K$$ the $$B(x,r) \subset K$$ for some $$r >0$$ Using the fact that $$x \in K$$ and $$K$$ is a subspace we get $$B(0,r) \subset K$$. If $$x$$ is any vector then there exits $$N$$ such that $$\frac x N \in B(0,r)$$. Hence $$x =N\frac x N \in K$$. Thus $$K=H$$.