# Finding a vector in a n.l.s.

Let $X$ be a normed linear space and $Y$ a closed proper subspace. Prove that for all $\varepsilon > 0$, there is an $x \in X$ with $\|x\| = 1$ and such that $\|x − y\| ≥ 1 − \varepsilon$ for all $y \in Y$ .

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Assume $\epsilon\in(0,1)$ (the result is trivial for $\epsilon\ge1$). Pick any $x$ in $X\setminus Y$. Then the distance $d$ from $x$ to $Y$ is positive (we use the hypothesis that $Y$ is closed here). Thus, you may choose a $z\in Y$ with $0\lt\Vert x-z\Vert <{d\over 1-\epsilon}$. Consider the vector $x-z\over\Vert x-z\Vert$.