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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
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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
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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
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1 Answer

  1. The definition of an "interval" given in the current edit trivially implies that every interval has empty interior in the cone of PSD matrices. The same holds for any other convex cone of dimension at least 2. Therefore, by the Baire category theorem, a countable union of such intervals cannot cover the cone.

  2. The alternative definition given in the edit, using entry-wise comparison of matrices, gives you "intervals" (I will call them "boxes") with nonempty interior. Then, trivially, there is countable cover of the cone by boxes (even a locally finite one, since the space is paracompact). I do not know if such a cover is "interesting", it depends on what you are interested in. You may want to define first the notion of an "interesting" cover of an open disk by rectangular boxes. On the other hand, since every closed box is compact, there is no finite covering (as the cone in question is noncompact).

  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.)

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Sorry, I must be dense (and so uncoverable...) but why in 2 is there e countable cover? –  Felix Goldberg Feb 26 '13 at 19:19
    
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
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