Let $Y$ be a proper subspace of $(X, \| \cdot \|)$. Is $\text{dist}(x,Y) > 0$ for $x \in X \setminus Y$?

Question

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?

Answer 1

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 $\inf_{y\in Y} \|x-y\| = 0$.

Answer 2

I just realised that the counterexample I came up with before asking this question is actually valid, in contrast to what I thought when constructing it.

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Consider $X := (L^1([0,1]), \| \cdot \|_1)$. Then $Y := \mathcal{C}([0,1])$ is a subset, which is not closed, as the sequnece $$ f_n \colon [0,1] \to [0,1], \ x \mapsto \begin{cases} 0, & x \in \left[0, \frac{1}{2} - \frac{1}{n}\right), \\ n x + 1 - \frac{n}{2}, & x \in \left[\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right], \\ 1, & x \in \left(\frac{1}{2}, 1 \right] \end{cases} $$ converges to $f(x) := \begin{cases} 0, & x \in \left[0, \frac{1}{2}\right), \\ 1, & x \in \left(\frac{1}{2}, 1 \right] \end{cases}$, which is not continuous.

We have $$ \inf_{y \in Y} \| f - y \|_1 \le \inf_{n \in \mathbb{N}} \| f - f_n \|_1 = \inf_{n \in \mathbb{N}} \frac{1}{2n} = 0. $$