Infinite Dimensional Vector Space: Finite Dim Subspace Closed and Nowhere Dense

Question

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.

Answer 1

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.

Answer 2

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.