Let $\mathcal H,\mathcal H'$ be Hilbert spaces (not necessarily separable).


A "separable state" is a trace-class operator of the form $\sum\_i \rho\_i\otimes\rho\_i'$ where $\rho\_i,\rho\_i'$ are positive trace-class operators over $\mathcal H,\mathcal H'$, respectively. (Convergence of the sum is with respect to the trace norm. $\otimes$ represents the tensor product.)


Is the set of separable states closed with respect to the trace norm?



Is the set of separable quantum states closed?
The answer is negative if we believe [1] (which comes without proofs, unfortunately):


[1] defines the set of separable states as the convex closure of $\{\sum\_{i=1}^n\sigma\_i\otimes\tau\_i\}$. I will call that set $S$. The set as defined in the original question I call $T$. Then $T\subseteq S$, and $T=S$ iff the answer is yes (i.e., iff $T$ is closed).


[1] shows that there is a $\rho\in S$ such that $\rho$ cannot be represented as a Bochner integral $\int\psi\psi^\*\otimes\phi\phi^\* \pi(d(\psi,\phi))$ for an atomic measure $\pi$. ($\rho$ is not "countably decomposable" in their language.) 


Any $\rho\in T$ can be represented as such an integral with a discrete and hence atomic measure $\pi$. Thus $T\subsetneq S$.




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On the positive side, if we define $T':=\{\int\psi\psi^\*\otimes\phi\phi^\* \pi(d(\psi,\phi))\}$ for probability measures $\pi$, then [1] shows $S=T'$. In particular, $T'$ is closed. So, in the spirit of the original question, the set of infinite convex combinations of $\sigma\_i\otimes\tau\_i$ is closed, only the notion of 
"infinite convex combination" must be changed: Not infinite sums, but integrals.




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[1] *Werner, R. F.; Kholevo, A. S.; Shirokov, M. E.*, [**On the concept of entanglement in Hilbert spaces.**](http://dx.doi.org/10.1070/RM2005v060n02ABEH000830), Russ. Math. Surv. 60, No. 2, 359-360 (2005); translation from Usp. Mat. Nauk 60, No. 2, 153-154 (2005). [ZBL1098.47019](https://zbmath.org/?q=an:1098.47019).


