Let $(X, \| \cdot \|)$ be an (infinite-dimensional) normed space, $Y \subset X$ a proper subspace and $x \in X \setminus Y$. Is $\text{dist}(x, Y) := \inf_{y \in Y} \| x – y \| > 0$?
My attempt.
Assume that $\text{dist}(x, Y) = 0$.
Then there exists a sequence $(y_n)_{n} \subset Y$ such that $\lim_{n \to \infty} \| x – y_n \| = 0$. Thus $y_n \to x$. If $Y$ were closed, then $x \in Y$, which would be a contradiction.
Is this statement true? If yes, how can I finish the proof?
Best Answer
The answer is no. For a counterexample, let $(X,\|\cdot\|)=(\ell^\infty,\|_\cdot\|_\infty)$ be the usual space of bounded sequences with the sup norm. Define $Y$ to be the subspace consisting of sequences that are eventually $0$ after finitely many terms. Then if we take $x=(1,\frac12,\frac13,\dots)\in X$ and $y_n = (1,\frac12,\dots,\frac1n,0,0,\dots)\in Y$, we have $\|x-y_n\| = \frac1{n+1}$ and therefore $\inf_{y\in Y} \|x-y\| = 0$.