Omission of the convexity hypothesis in the case of the projection theorem in a Hilbert space.

Question

I did not understand the following example concerning the omission of the convexity hypothesis in the case of the projection theorem in a Hilbert space.

let $Q$ the set of sequence $$x^{(n)}=\left(x_k^{(n)}\right)\in l^2$$ defined as $$x_k^{(n)}=0\quad\text{if}\quad k\ne n$$ $$x_k^{(n)}=1+\frac{1}{n}\quad\text{if}\quad k= n.$$

Then $Q$ is closed. As $$n\ne m\implies \lVert x^{(n)}-x^{(m)}\rVert>\sqrt {2},$$ $Q$ has no limit points in $l^2$.

Why can we conclude from this inequality that $Q$ is closed?

How can we prove that $ Q $ is not convex?

Answer 1

Why can we conclude from this inequality that $Q$ is closed?

This inequality implies that any Cauchy sequenece of elements of $Q$ has to be eventually constant. In particular convergent sequences in $Q$ are eventually constant, and so convergent to an element of $Q$.

More precisely, assume that $y^{(n)}$ is a sequence from $Q$, i.e. for any $n$ there is $m$ such that $y^{(n)}=x^{(m)}$. Now assume $y^{(n)}$ is Cauchy. Meaning for any $\epsilon>0$ there is $N$ such that for any $n,m\geq N$ the distance between $y^{(n)}$ and $y^{(m)}$ is at most $\epsilon$. Now if we take $\epsilon<\sqrt{2}$ then our inequality implies that $y^{(n)}=y^{(m)}$. Or in other words $y^{(n)}$ is eventually constant. In particular, $y^{(n)}$ converges to $y^{(N)}$, which is in $Q$.

How can we prove that $ Q $ is not convex?

Take for example $x^{(1)}$ and $x^{(2)}$ and their linear combination $\frac{1}{2}\cdot x^{(1)} + \frac{1}{2}\cdot x^{(2)}$. Note that this newly created sequence is non-zero at index $1$ and $2$ and thus it doesn't belong to $Q$.