# Covering the cone of positive semidefinite matrices by intervals

Is it possible to cover the cone of positive semidefinite matrices by a finite/countable/interesting family of closed intervals of matrices?

How about a general convex cone?

For the finite case the answer seems to be no but maybe there is some ingenious way I am missing.

EDIT: What I mean by interval $[A,B]$ is the set of convex combinations of the matrices $A$ and $B$. Other definitions of interval are possible, for example taking "convex combinations" with the weight scalars replaced by diagonal matrices. (like in this paper - http://www.math.wsu.edu/faculty/tsat/files/jt.pdf). Or ot can be defined entrywise, woth all matrices in the interval being entrywise greater than $A$ and less than $B$. Hope it's clearer now.

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What is an interval of matrices? (What is the partial order you are using?) Note that if you consider symmetric 2-by-2 positive definite matrices, the corresponding space is a positive cone of revolution, so covering it by finitely many rectangular boxes is out of question. –  Misha Feb 26 '13 at 0:34
By an interval do you mean a chain with respect to the usual ordering of PSD matrices? –  Yemon Choi Feb 26 '13 at 2:30
Misha: I'm afraid I don't follow what you mean in your last comment. By a chain I meant a totally ordered subset of a poset –  Yemon Choi Feb 26 '13 at 6:51
That said, I think this question is a bit too vague, or like a fishing expedition, in its present firm. Felix: is there any way to make more precise what you actually want to be true? –  Yemon Choi Feb 26 '13 at 6:53
With the clarification of the word "interval", for the finite/countable case, the answer is trivially no for measure-theoretic/dimension-theoretic reasons. For the "interesting" case it's hard to come up with an interpretation for which the answer is not trivially yes. –  Mark Meckes Feb 26 '13 at 14:43

3. With Yemon's interpretation, the answer is still the same as in (2), but one needs to do a bit more work. (Note that every convex cone $C$ in $R^n$ determines two partial orders on $R^n$: One where $C$ is positive and the other where the dual cone is positive, both definitions are natural.)
Felix: Every point in the cone is contained in the interior of a closed box $B\subset C$ (where $C$ is the open cone of positive-definite matrices). If you consider the closed cone of PSD matrices, then there is no countable covering by closed boxes (see the example of $2\times 2$ matrices): You cannot cover a closed round disk $D$ by countably many rectangular boxes contained in the disk. (Since every such box intersects the boundary circle in at most 4 points.) the case of $n\times n$ matrices reduces to the $2\times 2$ case. –  Misha Feb 26 '13 at 19:55