# If $U$ is a subspace of a normed space $X$, must points outside $U$ be a nonzero distance from $U$?

Let $$X$$ be a normed space over $$\mathbb K=\mathbb R$$ or $$\mathbb C$$ and $$U\subsetneq X$$ be a proper subspace. Suppose further that $$X\ni x \notin U$$.

1. Must we have $$d(x,U)>0$$ in general?
2. Must we have $$d(x,U)>0$$ if $$\dim U<\infty$$?

Note that $$d(x,U):=\inf_{u\in U} \| x-u \|$$.

I'm pretty sure the answer to (2) is "yes," since in this case $$X\simeq\mathbb K^{\dim U}$$, and then we can use topological theorems about $$\mathbb K^n$$ (as in this related question). Even so, I'd like to see a simpler proof that doesn't rely on the isomorphism with $$\mathbb K^n$$, since this seems like unnecessary baggage.

$$d(x,U)=0$$ if and only if $$x \in \overline U$$. So a dense porper subspace gives a counter-example to the first part. For a specific example take $$U=\ell^{0}$$ in $$X=\ell^{2}$$. Since finite dimensional subspaces are closed we have $$d(x,U) >0$$ when $$U$$ is finite dimensional.
1. take $$X$$ to be a space of convergent sequences and $$U$$ to be the subspace of $$X$$ comprising sequences that are finitely non-zero. This shows that claim (1) is false.
2. choose a basis $$\langle u_1, \ldots u_k \rangle$$ for $$U$$ and extend it to the basis $$\langle u_1, \ldots u_k, x \rangle$$ for $$\mathrm{span}(U, x)$$. If $$d(x, U) = 0$$, then $$x \in U$$. (This is essentially the argument that you don't like because of what you describe as excess baggage, but there is no way of avoiding it.)
• For (2), I still don't see how $d(x,U)=0\Rightarrow x\in U$ follows, unless we also use that $U$ is closed. Does this fact have an easy proof? Commented Nov 17, 2020 at 0:02
• Think about the ray $0x$ in $\mathrm{span}(U, x)$: how could $d(x, U)$ be non-zero if $x$ was not in $U$. Commented Nov 17, 2020 at 0:06