Some questions on convex sets. Are all bounded closed convex sets in a metric space $(M,d)$ compact? or if not are they complete?
The positive definite matrices form a convex set (Why does a positive definite matrix defines a convex cone?), do they also form a metric space? If so what is the metric? Are they complete or compact? 
P.S.
I am trying to understand if interior point methods can be applied to non Euclidean convex sets.
Thanks a lot.
 A: The set of positive definite matrices of a given size is convex. To see this, let $\mathbf A$ and $\mathbf B$ be two positive definite matrices in $\mathbb R^{n\times n}$, where $n$ is a positive integer. Let $\mathbf x\in\mathbb R^n$ be any nonzero vector and $\lambda\in[0,1]$. Then
\begin{align*}
\mathbf x^T\left[\lambda\mathbf A+(1-\lambda)\mathbf B\right]\mathbf x=\lambda \mathbf x^T\mathbf A\mathbf x+(1-\lambda)\mathbf x^T\mathbf B\mathbf x>0,
\end{align*}
since both quadratic forms are strictly positive (because $\mathbf A$ and $\mathbf B$ are positive definite and $\mathbf x$ is not zero) and at least one of $\lambda$ and $1-\lambda$ is positive.
For a given $n$, you can conceive the set of positive definite matrices of size $n$ as a subspace of $\mathbb R^{n^2}$ after rearranging the elements of the matrices into vectors. You can use the Euclidean metric on $\mathbb R^{n^2}$ on the subspace of vectors thus extracted from positive definite matrices. This set is not compact, because there exist positive definite matrices with arbitrarily large elements (say, $m\mathbf I$, where $m>0$ and $I$ is the identity matrix), nor is it complete, as $(1/m)\mathbf I$ is positive definite for all $m\in\mathbb Z_+$, but the limit of such a sequence is the zero matrix. (However, the set of positive semi-definite matrices, view as a subspace, is complete, because it is closed.)

To see a counterexample for a bounded, closed, convex set that is not compact, consider any infinite set $M$ that is also a vector space, endowed with the discrete metric $d$. That is, for any $x,y\in M$, let $$d(x,y)=\begin{cases} 0&\text{if $x=y$,}\\1&\text{otherwise.}\end{cases}$$ Then, $M$ is bounded (because the distance between any two points is at most $1$), convex (since $M$ is also a vector space), and closed (the whole space is always closed in any metric space), but $\{m\}_{m\in M}$ is an infinite collection of open sets (hint: in a discrete metric space, every set is open) whose union is $M$, but there is no finite subcollection whose union is still $M$ (because $M$ is infinite by assumption). Hence, $M$ is not compact. 
