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Show that any finite-dimensional subspace $(S,\|\cdot\|)$ of an infinite-dimensional normed vector space $(V,\|\cdot\|)$ is closed and nowhere dense.

Proof:

  • Let $\{x^{(n)}\}_{n\geq1}$ be a sequence in $S$. Suppose by contradiction that $x^{(n)}\to x \in V\backslash S$, infinite dimensional.
    By the definition of convergence $\|x^{(n)}-x\|\to 0$, but $x=(x_1,x_2,...)$ has at least one coordinate such that $x_i\neq 0= x_i^{(n)}$ for some $i=1,2,...$, where $x_i^{(n)}$ is the $i$-th element of the $n$-th vector of the sequence, and so a contradiction.
  • Not sure what topology should I refer to (product top?!)

Question: can anyboody check the firsts part and give a hint for the second.

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    $\begingroup$ Why is it a contradiction if some element of $x$ is non zero? What do you mean by coordinate? Does $V$ have some structure in addition to being a normed space? The topology is surely the topology that comes with the associated norm? $\endgroup$ – copper.hat Feb 23 '14 at 0:36
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  1. The starting idea to take a sequence $x_n$ in $S$ that converges to some $x$ and see whether $x\in S$ is OK, the rest is nonsense, e.g. nobody said that $x=0$ neither that $x_i^{(n)}=0$, or $S$ may have more than countably infinite dimensions, whence the sequence setting $x=(x_1,x_2,\dots)$ is not that much justified.
    Probably the fastest argument to continue is to observe that $S$ is complete, i.e. is a Banach space (because finite dimensional), and then, since $x_n$ converges to $x\in V$, we have that $x_n$ is a Cauchy sequence, but then it also converges in $S$, so we must have $x\in S$.

  2. The topology is given by the norm: open balls $B_\varepsilon(x):=\{v\in V\mid \|v-x\|<\varepsilon\}$ form a basis of the topology, just as in the finite dimensional case. Hence, every open set is a union of open balls, but any open ball 'uses up' all the dimensions.

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Suppose $p \notin S$, and let $v_1,...v_n$ span $S$. Define the functional $f$ on the space spanned by $p,v_1,...v_n$ by $f(v_k) = 0$, $f(p) = 1$. Use Hahn Banach to extend this functional to the whole space. Let $U = \{ x | f(x) > {1 \over 2} \}$ which is open, and since $f(s) = 0$ when $s \in S$, we see that $S \cap U = \emptyset$, hence $S^c$ is open.

Note: The point of the above construction is to create a continuous functional $f$ such that $S \subset \ker f$ and $p \notin \ker f$. Then the set $\{x | f(x) >0 \}= f^{-1}((0,\infty))$ is open (since $f$ is continuous) and $\ker f \cap f^{-1}((0,\infty)) = \emptyset$.

By the same token, we see that if $f(x) \neq 0$, then $x \notin S$.

To see that $S$ is nowhere dense, note that since $S$ is closed, we need only show that it has an empty interior. Suppose $s \in S$, then let $x_n = s+{1 \over n} p$ (where $p$ is from the first paragraph).

Spoiler:

Note that $f(x_n) = {1 \over n}$ and so $x_n \notin S$, and we have $\|s-x_n\| = {1 \over n}\|p\|$. Hence $S$ contains no open set.

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  • $\begingroup$ Thank you for the very good answer. I though shamefully cannot get the bit "$S \cap U = \emptyset$, hence $S^c$ is open". Is it because somehow if $x\not\in S$ then $x\in U$ by the definition of the linear bdd functional? Could not it be that there is another element $z$ in $S^c$ s.t. $f(z)=0 $? $\endgroup$ – Ton Feb 23 '14 at 16:46
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    $\begingroup$ Showing that $S$ is closed is equivalent to showing that $S^c$ is open. So I picked a point $p \in S^c$ and created an open set containing $p$ that does not intersect $S$, which shows that $S^c$ is open. It is possible that $\ker f$ contains points other than $S$, but it doesn't matter for the purpose of showing that $S^c$ is open. $\endgroup$ – copper.hat Feb 23 '14 at 17:03

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