[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. The implication $(1) \Rightarrow (2)$ is immediate from the [Hierarchy of Operator Topologies](/theorems/1237). For $(3) \Rightarrow (1)$, we show that the commutant $\mathcal{S}'$ of any subset $\mathcal{S} \subset \mathcal{L}(H, H)$ is WOT-closed, so $\mathcal{M}'' = (\mathcal{M}')'$ is WOT-closed, and $\mathcal{M} = \mathcal{M}''$ inherits this property. The hard direction $(2) \Rightarrow (3)$ proceeds as follows: given $T \in \mathcal{M}''$, we show $T$ lies in the SOT-closure of $\mathcal{M}$ by a finite-dimensional approximation. For any $x_1, \ldots, x_n \in H$ and $\varepsilon > 0$, we construct $S \in \mathcal{M}$ with $\|Sx_i - Tx_i\| < \varepsilon$ for all $i$. The construction uses the fact that $\mathcal{M}$-invariant subspaces are also $\mathcal{M}''$-invariant (via the orthogonal projection argument), reducing the problem to a single-vector approximation iterated on the direct sum $H^n$. [/proofplan]

[step:Show $(1) \Rightarrow (2)$: WOT-closed implies SOT-closed] By the [Hierarchy of Operator Topologies](/theorems/1237), every SOT-open set is WOT-open. Equivalently, every WOT-closed set is SOT-closed. Since $\mathcal{M}$ is WOT-closed by hypothesis, $\mathcal{M}$ is SOT-closed. [/step]

[step:Show $(3) \Rightarrow (1)$: $\mathcal{M} = \mathcal{M}''$ implies WOT-closed]It suffices to show that the commutant $\mathcal{S}'$ of any subset $\mathcal{S} \subset \mathcal{L}(H, H)$ is WOT-closed. Let $(T_\alpha)$ be a net in $\mathcal{S}'$ with $T_\alpha \to T$ in the WOT. For any $A \in \mathcal{S}$ and any $x, y \in H$: \begin{align*} (TAx, y)_H &= \lim_\alpha (T_\alpha A x, y)_H = \lim_\alpha (A T_\alpha x, y)_H, \end{align*} where the second equality uses $T_\alpha \in \mathcal{S}'$, i.e., $T_\alpha A = A T_\alpha$. Now $(AT_\alpha x, y)_H = (T_\alpha x, A^* y)_H \to (Tx, A^* y)_H = (ATx, y)_H$ by the WOT convergence of $T_\alpha$ (applied to the vector $x$ and the functional $(\cdot, A^* y)_H \in H^*$). Therefore $(TAx, y)_H = (ATx, y)_H$ for all $x, y \in H$, and since the inner product separates points, $TA = AT$. Since $A \in \mathcal{S}$ was arbitrary, $T \in \mathcal{S}'$. Therefore $\mathcal{S}'$ is WOT-closed. In particular, $\mathcal{M}' = \mathcal{M}'$ is WOT-closed, and $\mathcal{M}'' = (\mathcal{M}')'$ is WOT-closed. If $\mathcal{M} = \mathcal{M}''$, then $\mathcal{M}$ is WOT-closed.[/step]

[guided]We show that commutants are always WOT-closed, regardless of any structural assumptions on $\mathcal{S}$. The commutant $\mathcal{S}' = \{T \in \mathcal{L}(H,H) : TA = AT \text{ for all } A \in \mathcal{S}\}$ can be written as an intersection: \begin{align*} \mathcal{S}' = \bigcap_{A \in \mathcal{S}} \{T \in \mathcal{L}(H,H) : TA = AT\}. \end{align*} An intersection of closed sets is closed, so it suffices to show each set $\{T : TA = AT\}$ is WOT-closed. Fix $A$ and let $T_\alpha \to T$ in the WOT with $T_\alpha A = AT_\alpha$ for all $\alpha$. For any $x, y \in H$: \begin{align*} (TAx, y)_H = \lim_\alpha (T_\alpha Ax, y)_H = \lim_\alpha (AT_\alpha x, y)_H = \lim_\alpha (T_\alpha x, A^*y)_H = (Tx, A^*y)_H = (ATx, y)_H. \end{align*} The first and fourth equalities use WOT convergence (with pairs $(Ax, y)$ and $(x, A^*y)$ respectively). The second equality uses $T_\alpha A = AT_\alpha$. Since $(TAx, y)_H = (ATx, y)_H$ for all $x, y$, we conclude $TA = AT$, so $T \in \mathcal{S}'$. The *-subalgebra and unital hypotheses play no role here — commutants of arbitrary sets are WOT-closed. These hypotheses are consumed in the hard direction $(2) \Rightarrow (3)$.[/guided]

[step:Establish that $\mathcal{M}$-invariant closed subspaces are $\mathcal{M}''$-invariant][claim:Invariant subspace lemma] Let $V \subset H$ be a closed subspace that is $\mathcal{M}$-invariant (i.e., $Av \in V$ for all $A \in \mathcal{M}$, $v \in V$). Then $V$ is $\mathcal{M}''$-invariant: for every $T \in \mathcal{M}''$ and $v \in V$, $Tv \in V$. [/claim] [proof] Let $P: H \to H$ be the orthogonal projection onto $V$. We show $P \in \mathcal{M}'$, which will give the result. Since $\mathcal{M}$ is a *-subalgebra, $V$ is invariant under both $A$ and $A^*$ for every $A \in \mathcal{M}$. To see this: for $A \in \mathcal{M}$, $V$ is $A$-invariant by hypothesis. For $A^*$-invariance of $V$: let $v \in V$ and $w \in V^\perp$. Then $(A^* v, w)_H = (v, Aw)_H$. Since $\mathcal{M}$ is a *-algebra, $A^* \in \mathcal{M}$, and $V$ is $\mathcal{M}$-invariant, so $V$ is also $A^*$-invariant. Since $V$ is invariant under every $A \in \mathcal{M}$, the orthogonal complement $V^\perp$ is invariant under $A^*$ for each $A \in \mathcal{M}$: for $w \in V^\perp$ and $v \in V$, $(A^* w, v)_H = (w, Av)_H = 0$ since $Av \in V$. But since $\mathcal{M}$ is closed under adjoints, $A^* \in \mathcal{M}$, so $V^\perp$ is invariant under all operators in $\mathcal{M}$. Now for any $A \in \mathcal{M}$ and $x \in H$, write $x = Px + (I - P)x$ with $Px \in V$ and $(I-P)x \in V^\perp$. Then $APx \in V$ and $A(I-P)x \in V^\perp$ (by $\mathcal{M}$-invariance of both $V$ and $V^\perp$). Therefore: \begin{align*} PAx = P(APx + A(I-P)x) = APx + 0 = APx. \end{align*} Hence $PA = AP$ for all $A \in \mathcal{M}$, so $P \in \mathcal{M}'$. Now let $T \in \mathcal{M}''$. Since $P \in \mathcal{M}'$, $TP = PT$. For any $v \in V$: $Tv = T(Pv) = P(Tv) \in V$. Therefore $V$ is $\mathcal{M}''$-invariant. [/proof][/step]

[guided]The invariant subspace lemma is the geometric core of the Double Commutant Theorem. It says that the algebraic condition $T \in \mathcal{M}''$ (commuting with everything that commutes with $\mathcal{M}$) forces $T$ to respect all closed invariant subspaces of $\mathcal{M}$. Why does the *-subalgebra hypothesis matter? If $\mathcal{M}$ were not closed under adjoints, a closed $\mathcal{M}$-invariant subspace $V$ need not have an $\mathcal{M}$-invariant complement. Consider a non-self-adjoint operator $A$ on $\mathbb{C}^2$: $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has invariant subspace $V = \operatorname{span}\{e_1\}$, but $A^* = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ does not leave $V$ invariant, so $V^\perp = \operatorname{span}\{e_2\}$ is not $A$-invariant. Without invariance of $V^\perp$, the projection $P$ onto $V$ does not commute with $A$. The *-algebra hypothesis ensures that whenever $V$ is $\mathcal{M}$-invariant, $V^\perp$ is also $\mathcal{M}$-invariant (since $A^* \in \mathcal{M}$ and $V^\perp$ is $A^*$-invariant whenever $V$ is $A$-invariant). This makes the orthogonal projection $P$ onto $V$ commute with all of $\mathcal{M}$, placing $P$ in $\mathcal{M}'$, and any $T \in \mathcal{M}''$ must then commute with $P$ and hence preserve $V$.[/guided]

[step:Prove the single-vector approximation: for $T \in \mathcal{M}''$ and $x \in H$, $Tx \in \overline{\mathcal{M}x}$]Fix $T \in \mathcal{M}''$ and $x \in H$. Define $V := \overline{\mathcal{M}x} = \overline{\{Ax : A \in \mathcal{M}\}}$, the closure of the $\mathcal{M}$-orbit of $x$. $V$ is a closed subspace of $H$: it is the closure of the linear span $\mathcal{M}x$. This span is a linear subspace since $\mathcal{M}$ is a linear subspace of $\mathcal{L}(H,H)$ (being a subalgebra): for $A, B \in \mathcal{M}$ and $\lambda \in \mathbb{R}$, $(\lambda A + B)x = \lambda Ax + Bx \in \mathcal{M}x$. $V$ is $\mathcal{M}$-invariant: for $A \in \mathcal{M}$ and $Bx \in \mathcal{M}x$, $A(Bx) = (AB)x \in \mathcal{M}x$ since $AB \in \mathcal{M}$. By continuity of $A$ and density of $\mathcal{M}x$ in $V$, $A(V) \subset V$. By the invariant subspace lemma, $V$ is $\mathcal{M}''$-invariant. Since $T \in \mathcal{M}''$, $Tx \in V$. Moreover, $x \in V$ because $I \in \mathcal{M}$ (the unital hypothesis), so $Ix = x \in \mathcal{M}x \subset V$. Therefore $Tx \in V = \overline{\mathcal{M}x}$: for every $\varepsilon > 0$, there exists $A \in \mathcal{M}$ with $\|Ax - Tx\|_H < \varepsilon$.[/step]

[guided]This step reduces the Double Commutant Theorem to a geometric statement. Given $T \in \mathcal{M}''$ and a single vector $x$, we must find operators in $\mathcal{M}$ that approximate $Tx$. The natural candidate is the orbit $\mathcal{M}x = \{Ax : A \in \mathcal{M}\}$. Its closure $V = \overline{\mathcal{M}x}$ is a closed subspace (using the linearity of $\mathcal{M}$). It is $\mathcal{M}$-invariant because $\mathcal{M}$ is closed under multiplication: $A(Bx) = (AB)x \in \mathcal{M}x$. The unital hypothesis $I \in \mathcal{M}$ ensures $x = Ix \in \mathcal{M}x \subset V$, so the orbit is non-degenerate. By the invariant subspace lemma (which uses the *-subalgebra hypothesis), $V$ is $\mathcal{M}''$-invariant. Since $T \in \mathcal{M}''$, $Tx \in V = \overline{\mathcal{M}x}$. This means $Tx$ can be approximated arbitrarily well by elements of $\mathcal{M}x$ — i.e., for any $\varepsilon > 0$, there exists $A \in \mathcal{M}$ with $\|Ax - Tx\| < \varepsilon$. The difficulty is that the approximating operator $A$ depends on $x$. For SOT convergence, we need a *single* operator that simultaneously approximates $Tx_1, \ldots, Tx_n$ for finitely many vectors. This requires the amplification trick of the next step.[/guided]

[step:Amplify to simultaneous approximation on $H^n$ and conclude $(2) \Rightarrow (3)$]Assume $\mathcal{M}$ is SOT-closed. We must show $\mathcal{M}'' \subset \mathcal{M}$. Fix $T \in \mathcal{M}''$, vectors $x_1, \ldots, x_n \in H$, and $\varepsilon > 0$. We must find $S \in \mathcal{M}$ with $\|Sx_i - Tx_i\|_H < \varepsilon$ for all $i = 1, \ldots, n$ (this is the condition for $T$ to lie in the SOT-closure of $\mathcal{M}$). Consider the direct sum Hilbert space $H^n = H \oplus \cdots \oplus H$ ($n$ copies) with inner product $((u_1, \ldots, u_n), (v_1, \ldots, v_n))_{H^n} = \sum_{i=1}^n (u_i, v_i)_H$. For $A \in \mathcal{L}(H, H)$, define the amplification \begin{align*} A^{(n)}: H^n &\to H^n \\ (u_1, \ldots, u_n) &\mapsto (Au_1, \ldots, Au_n). \end{align*} Define $\widetilde{\mathcal{M}} := \{A^{(n)} : A \in \mathcal{M}\} \subset \mathcal{L}(H^n, H^n)$. [claim:The double commutant of $\widetilde{\mathcal{M}}$ contains $T^{(n)}$] $T^{(n)} \in \widetilde{\mathcal{M}}''$. [/claim] [proof] We first compute $\widetilde{\mathcal{M}}'$. An operator $R \in \mathcal{L}(H^n, H^n)$ can be represented as an $n \times n$ matrix of operators $(R_{ij})_{1 \le i,j \le n}$ with $R_{ij} \in \mathcal{L}(H, H)$, where $(Ru)_i = \sum_{j=1}^n R_{ij} u_j$. The condition $R A^{(n)} = A^{(n)} R$ for all $A \in \mathcal{M}$ reads: \begin{align*} \sum_{j=1}^n R_{ij} A u_j = A \sum_{j=1}^n R_{ij} u_j \quad \text{for all } u \in H^n, \; A \in \mathcal{M}, \; i = 1, \ldots, n. \end{align*} Choosing $u = e_j \otimes v$ (i.e., $v$ in the $j$-th slot, zero elsewhere), this becomes $R_{ij} A v = A R_{ij} v$ for all $v \in H$, i.e., $R_{ij} \in \mathcal{M}'$. Conversely, any matrix $(R_{ij})$ with $R_{ij} \in \mathcal{M}'$ commutes with all $A^{(n)}$. Therefore \begin{align*} \widetilde{\mathcal{M}}' = \{(R_{ij})_{1 \le i,j \le n} : R_{ij} \in \mathcal{M}' \text{ for all } i, j\} = M_n(\mathcal{M}'). \end{align*} Now compute $\widetilde{\mathcal{M}}''$. An operator $Q \in \mathcal{L}(H^n, H^n)$ with matrix $(Q_{ij})$ lies in $\widetilde{\mathcal{M}}'' = (\widetilde{\mathcal{M}}')'$ if and only if it commutes with every $R = (R_{ij}) \in M_n(\mathcal{M}')$. In particular, taking $R = E_{ij} \otimes B$ (the matrix with $B \in \mathcal{M}'$ in position $(i,j)$ and zero elsewhere), the condition $QR = RQ$ gives $Q_{ki} B = B Q_{kj}$ for $k \ne i$ and $k \ne j$ by examining the appropriate entries. Taking $B = I \in \mathcal{M}'$ (since $I$ commutes with everything) and $R = E_{ij} \otimes I$ (the permutation-like matrix), the commutation condition forces $Q$ to be a diagonal matrix: $Q_{ij} = 0$ for $i \ne j$, and $Q_{11} = Q_{22} = \cdots = Q_{nn}$. Setting $Q_0 := Q_{11}$, the condition further requires $Q_0 B = B Q_0$ for all $B \in \mathcal{M}'$, i.e., $Q_0 \in \mathcal{M}''$. Therefore $\widetilde{\mathcal{M}}'' = \{Q_0^{(n)} : Q_0 \in \mathcal{M}''\}$. Since $T \in \mathcal{M}''$, we have $T^{(n)} \in \widetilde{\mathcal{M}}''$. [/proof] Now apply the single-vector approximation result to $\widetilde{\mathcal{M}}$ acting on $H^n$ with $T^{(n)} \in \widetilde{\mathcal{M}}''$ and the vector $\mathbf{x} := (x_1, \ldots, x_n) \in H^n$. We must verify that $\widetilde{\mathcal{M}}$ is a unital *-subalgebra of $\mathcal{L}(H^n, H^n)$: it is unital since $I^{(n)} = I_{H^n}$; it is closed under multiplication since $(AB)^{(n)} = A^{(n)} B^{(n)}$; and it is closed under adjoints since $(A^{(n)})^* = (A^*)^{(n)}$ and $A^* \in \mathcal{M}$. By the single-vector approximation, $T^{(n)} \mathbf{x} \in \overline{\widetilde{\mathcal{M}} \mathbf{x}}$. Therefore there exists $S \in \mathcal{M}$ with \begin{align*} \|S^{(n)} \mathbf{x} - T^{(n)} \mathbf{x}\|_{H^n} < \varepsilon, \end{align*} i.e., $\sum_{i=1}^n \|Sx_i - Tx_i\|_H^2 < \varepsilon^2$. In particular, $\|Sx_i - Tx_i\|_H < \varepsilon$ for each $i = 1, \ldots, n$. Since $x_1, \ldots, x_n$ and $\varepsilon$ were arbitrary, $T$ lies in the SOT-closure of $\mathcal{M}$. Since $\mathcal{M}$ is SOT-closed, $T \in \mathcal{M}$. Since $T \in \mathcal{M}''$ was arbitrary, $\mathcal{M}'' \subset \mathcal{M}$. The reverse inclusion $\mathcal{M} \subset \mathcal{M}''$ holds for any set (every operator commutes with everything that commutes with it), so $\mathcal{M} = \mathcal{M}''$.[/step]

[guided]The single-vector approximation shows that for each individual $x$, there exists $A \in \mathcal{M}$ with $Ax \approx Tx$. But the SOT topology requires simultaneous approximation at finitely many vectors: we need a *single* $S \in \mathcal{M}$ with $Sx_i \approx Tx_i$ for $i = 1, \ldots, n$. The operator $A$ from the single-vector step depends on $x$, so we cannot directly reuse it. The amplification trick resolves this by passing to the direct sum $H^n$. The key insight is that if $T \in \mathcal{M}''$, then the diagonal operator $T^{(n)}$ lies in $\widetilde{\mathcal{M}}''$, and the single-vector approximation on $H^n$ with the vector $\mathbf{x} = (x_1, \ldots, x_n)$ produces $S^{(n)} \in \widetilde{\mathcal{M}}$ with $\|S^{(n)}\mathbf{x} - T^{(n)}\mathbf{x}\|_{H^n} < \varepsilon$. Since $S^{(n)}$ acts diagonally as $S$ on each component, this single operator $S$ simultaneously approximates $Tx_i$ at every $x_i$. The computation of $\widetilde{\mathcal{M}}'$ and $\widetilde{\mathcal{M}}''$ is the technical heart. The commutant $\widetilde{\mathcal{M}}'$ consists of all $n \times n$ matrices over $\mathcal{M}'$: commuting with all diagonal operators $A^{(n)}$ forces each entry $R_{ij}$ to commute with every $A \in \mathcal{M}$, but places no constraint on how different entries relate. The double commutant $\widetilde{\mathcal{M}}''$ is much more rigid: commuting with *all* matrices over $\mathcal{M}'$ — including the "off-diagonal" ones $E_{ij} \otimes B$ — forces the operator to be scalar-diagonal: $Q = Q_0^{(n)}$ with $Q_0 \in \mathcal{M}''$. To verify this last claim more explicitly: take $R = E_{ij} \otimes I$ (the operator that maps the $j$-th component to the $i$-th component and annihilates all others). If $Q = (Q_{kl})$ commutes with $R$, then $(QR)_{ki} = Q_{kj}$ while $(RQ)_{ki} = Q_{ii}$ when $k = i$ and $0$ when $k \ne i$... Let us be more careful. $(RQ)_{kl} = \delta_{ki} Q_{jl}$ and $(QR)_{kl} = Q_{ki} \delta_{jl}$. Setting these equal for all $k, l$: for $l = j$, $\delta_{ki} Q_{jj} = Q_{ki}$, so $Q_{ki} = 0$ for $k \ne i$ and $Q_{ii} = Q_{jj}$. Since this holds for all $i, j$, the matrix $Q$ is a scalar diagonal $Q_0 I_n$ with $Q_0 := Q_{11} = \cdots = Q_{nn}$. The further requirement $Q_0 B = BQ_0$ for all $B \in \mathcal{M}'$ (from commuting with $E_{11} \otimes B$) gives $Q_0 \in (\mathcal{M}')' = \mathcal{M}''$.[/guided]