If a <linear subspace> $V$ of a <normed space> contains an open ball $B(v,r)$, then $v\in V$ and subtraction gives $B(0,r)\subseteq V$. For any $x\in X$, choose a positive scalar $t$ with $\|tx\|<r$. Then $tx\in V$ and hence $x\in V$. Thus $V=X$. \b[Every proper linear subspace has empty interior.]

Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 22H / b / Solution

 Solution (source code)