TITLE: Every proper subspace of a normed linear space is not open
QUESTION [6 upvotes]: Possible Duplicate:
Interior of a Subspace 

How do we show that any proper subspace of a normed linear space is not open. I know that for nay finite dimensional normed linear space $(X,||.||)$ any proper subspace is closed but I am not sure how to show this?
Thanks for any help

REPLY [14 votes]: It is enough to show that the subspace does not contain any open ball centered on $0$. Assume we have such a ball $B_\delta(0)$.
Since the subspace is proper we can choose some vector $v$ outside the subspace. It is nonzero because $0$ is in every subspace, so its norm is nonzero too. Therefore by appropriate scaling there is a scalar $\lambda$ such that $\|\lambda v\|=\delta/2$. Can $\lambda v$ be in the subspace?

REPLY [4 votes]: Hint: 
If $Y$ is an open subspace of $X$, then $0$ is its inner point, so $Y$ contains one of its neighborhoods, say $B_\varepsilon(0)=\{x\mid ||x||<\varepsilon\}$.