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.
Best Answer
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 $x\in S$.
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.