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Use the conic Carathéodory theorem in the real space of symmetric matrices, of dimension $N=n(n+1)/2$. A convergent sequence $B_{\ell }$ in the completely positive cone has padded factorizations $B_{\ell }={\sum }_{j=1}^N{x}_{\ell j}{x}_{\ell j}^T$ with nonnegative factors. Their total squared norms equal $\mathrm{tr}{B}_{\ell }$ and are uniformly bounded. A simultaneous convergent subsequence of the finite factor tuple gives $B={\sum }_j{x}_j{x}_j^T$ with $x_j\ge 0$. Thus the limit remains in the cone. The argument also bounds the required number of factors by $N$.

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.