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The convex cone of completely positive matrices:It is a closed convex cone and the dual cone of the copositive cone under the Frobenius inner product. The finite-sum definition imposes no closure by fiat; closedness of the completely positive cone supplies that fact.

Under the nonnegative-pairing convention for the dual cone, ${\mathrm{CP}}_n^{\ast }={\mathrm{COP}}_n$ follows from $⟨A,x{x}^T{⟩}_F=x^{T}Ax$. Conversely, if $B\notin {\mathrm{CP}}_n$, separation from a closed convex cone gives a separating $A$ nonnegative on all generators and negative on $B$. That $A$ is a copositive matrix, so $B\notin {\mathrm{COP}}_n^{\ast }$. This proves the displayed equality using closedness of the completely positive cone.

The real vector space of symmetric matrices has dimension $N=n(n+1)/2$. By the conic Carathéodory theorem, every member of $C_0$ is a conic combination of at most $N$ generators. Absorb each nonnegative coefficient into its vector through $\lambda x{x}^T=(\sqrt{\lambda }x)(\sqrt{\lambda }x{)}^T$.

If $B_{\ell }\in C_0$ converges to $B$, write, padding with zero vectors if necessary,The matrix trace satisfiesThe left side is bounded because $B_{\ell }$ converges. Thus the finite tuple $(x_{\ell 1},\dots ,x_{\ell N})$ is bounded. The Bolzano-Weierstrass theorem gives a subsequence on which every vector converges, say $x_{\ell j}\to x_j\ge 0$. Continuity of the outer product now gives $B={\sum }_{j=1}^N{x}_j{x}_j^T\in C_0$. HenceThis proves closedness of the completely positive cone. The uniform bound on the number of factors and the matrix trace bound are both essential: an arbitrary conic hull of a closed generating set need not be closed.