Schroeder-Bernstein theorem for representations of C*-algebras I am trying to work on an exercise which claims that
If two representations ρ and σ, on Hilbert spaces H and G respectively, are each unitarily equivalent to a subrepresentation of the other, then they are unitarily equivalent, which is called the
Schroeder-Bernstein theorem for representations of C*-algebras.
You can refer to
https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorems_for_operator_algebras
for more details about the theorem and an outline of the proof.
I am now confused about the outline he gives in the wiki, where he claims that by induction we can get $\rho_1\simeq (\bigoplus_{i\geq1}\rho_i^\prime)\bigoplus (\bigoplus_{i\geq1}\sigma_i^\prime)$. I am not sure we can get this by induction. What we do get is that $\rho_1\simeq (\bigoplus_{i=1}^{n}\rho_i^\prime)\bigoplus 1(\bigoplus_{i=1}^{n}\sigma_i^\prime)\bigoplus \rho_{n+1}$ for every $n\in\mathbb{N}$, where $\rho_1\simeq \rho_{n+1}$, but this doen't mean that we can claim the equation for countable infinity.
Any help will be truly grateful!
 A: The section "Representations of C$^*$-algebras" is poorly written, but the idea of what needs to be  done  is in the previous section.
To work as in the subspace case, we first note that a required hypothesis is that both representations are non-degenerate (otherwise, the result is not true, as we can consider cases like $\rho$ and $\sigma=\rho\oplus0$ and there is no unitary between the ambient Hilbert spaces). It follows that $\dim H=\dim G$, and so we may assume without loss of generality (at the cost of one more unitary conjugation) that $G=H$.
By hypothesis $\rho=\sigma_1\oplus\rho_1'$, with $\sigma_1\simeq\sigma$. This means that there exists a subspace $N_1$ with  $\rho|_{N_1}=\sigma_1$. Let $R_1=H\ominus N_1$, so $\rho|_{R_1}=\rho_1'$. Next we have $\sigma_1=\rho_2\oplus\sigma_2'$, where $\rho_2\simeq\rho$. So there exists a subspace $M_1\subset N_1$ with  $\sigma_1|_{M_1}=\rho_2$. Let $S_1=N_1\ominus M_1$, so that $\sigma_1|_{S_1}=\sigma_2'$. By induction we get subspaces and representations
$$
\rho_k=\sigma_k\oplus\rho_k',\qquad \sigma_k=\rho_{k+1}\oplus\sigma_{k+1}',
$$
$$
\rho_k|_{N_k}=\sigma_k,\quad \rho_k|_{R_k}=\rho_k',
$$
$$
\sigma_k|_{M_k}=\rho_k,\quad \sigma_k|_{S_k}=\sigma_{k+1}',
$$
$$
R_j=M_{j-1}\ominus N_j,\qquad S_j=N_j\ominus M_j
$$
and
$$
H\supset N_1\supset M_1\supset N_2\supset M_2\supset\cdots
$$
Since each inclusion $N_{j+1}\subset M_j$ and $M_j\subset N_j$ arises from the same decomposition, we have $\rho_j'\simeq\rho_k'$, $\sigma_j'\simeq\sigma_k'$ for all $k,j$. Here's a poor man's picture of the inclusions.

If we let $Q=\bigcap_jM_j=\bigcap_jN_j$, we can write
$$
H=Q\oplus\,\bigoplus_{k=1}^\infty R_k\,\oplus\,\bigoplus_{k=1}^\infty S_k,\qquad N_1=Q\oplus\,\bigoplus_{k=2}^\infty R_k\,\oplus\,\bigoplus_{k=1}^\infty S_k.
$$
We have
$$
\rho|_{R_k}=\rho_k',\qquad \rho|_{S_k}=\sigma_{k+1}',\qquad \sigma_1|_{R_{k+1}}=\rho_{k+1}',\qquad \sigma_1|_{S_k}=\sigma_{k+1}',
$$
and, very importantly,  $\rho|_Q=\sigma_1|_Q$. Let $\gamma=\rho|_Q=\sigma|_Q$. The decomposition then gives us
$$
\rho=\gamma\oplus\,\bigoplus_{k=1}^\infty \rho_k'\,\oplus\,\bigoplus_{k=1}^\infty \sigma_{k+1}',\qquad \sigma_1=\gamma\oplus\,\bigoplus_{k=2}^\infty \rho_k'\,\oplus\,\bigoplus_{k=1}^\infty \sigma_{k+1}'.
$$
Now one can generate a unitary by shifting the index in the sum of the $\rho'$ to get $\rho\simeq \sigma_1$.
