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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.

WillG
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  • Related posts: https://math.stackexchange.com/questions/1883094/every-finite-dimension-subspace-of-mathcalx-is-closed (answers your first question) and https://math.stackexchange.com/questions/2066765/subspaces-of-ell2-and-ell-infty-which-are-not-closed (answers your second question in the negative when $X = \ell^2$) – aras Nov 16 '20 at 23:53

2 Answers2

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Hints:

  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.)

Rob Arthan
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  • 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? – WillG Nov 17 '20 at 00:02
  • Think about the ray $0x$ in $\mathrm{span}(U, x)$: how could $d(x, U)$ be non-zero if $x$ was not in $U$. – Rob Arthan Nov 17 '20 at 00:06
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$d(x,U)=0$ 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.

Kavi Rama Murthy
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