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?


2 Answers 2


No, you’re right, an open ball always linearly generates the full space :

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.

Therefore a subspace that contains a ball is the entire space.

Therefore all strict subspace have empty interior


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 $K=H$.


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