Orthonormal Sets and Compactness 1) Let $\{u_n\}$ $(n=1,2,\ldots)$ be an orthonormal set in Hilbert space $H$. Show that this set is closed and bounded but not compact. 
2) Let $Q$ be the set of all $x\in H$ of the form $$x=\sum\limits_{n = 1}^\infty c_n u_n$$ where $|c_n|\leq1/n$. Prove $Q$ is compact.
3) More generally, let $\{\delta_n\}$ be a sequence of positive numbers, and let $S$ be the set of all $x\in H$ of the form $$x=\sum_1^\infty c_n u_n$$
where $|c_n| \leq \delta_n$. Prove that $S$ is compact iff $$\sum_1^\infty\delta_n^2<\infty$$
4) Prove that $H$ is not locally compact.
I don't really even know where to start with this.
 A: Boundedness of $\{u_n\}$ is straightforward.
To show it is closed, show the complement is open (expand $\{u_n\}$ to a basis, fix $x\not\in \{u_n\}$, write $x$ in terms of your basis, note the coefficients of $x$ in this basis must decay, then show $\|x - u_n\| \geq \epsilon$ for some $\epsilon$ independent of $n$).
Let $B_{1/2}(u_n) = \{u \in H : \|u - u_n\| < 1/2\}$. Then $\bigcup_n B_{1/2}(u_n) $ is an open cover of $\{u_n\}$. Is there any finite subcover (hint: compute $\|u_n - u_m\|$ for $n\neq m$ using orthogonality)?
A: The distance between any two of these basis elements is $\|u_i - u_j\|=\sqrt 2$.  Consequently if you put an open ball of radius $<\sqrt 2/2$ around each of them, those balls are pairwise disjoint.  That means any sequence of these points other than $u_i$ must stay out of that ball centered at $u_i$; hence $u_i$ is not a limit point of the set.  Since no point of the set is a limit point of the set, the set is closed.
Hence it is closed and bounded.
But the same set of open balls described above is an open cover with no finite subcover.  (Indeed, it has no proper subcover at all: if you omit even one of those open balls, the remaining ones fail to cover the set.)  Therefore the set is not compact.
If $H$ were locally compact, then every open set would have compact closure.  That would imply every closed ball is compact.  Notice that the argument above shows the closed ball of radius $1$ centered at $0$ cannot be compact because it has a closed subset that is not compact.  Now observe that every closed ball is homeomorphic to that one.  Hence $H$ is not locally compact.
