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Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...

3

The result is that $\varphi_\tau(M)$ is a von Neumann algebra if and only if $\tau$ is normal. If $\tau$ is normal, then so is $\varphi_\tau$ and so $\varphi_\tau(M)$ is a von Neumann algebra. Conversely, suppose that $\varphi_\tau(M)$ is a von Neumann algebra. Let $\{x_j\}$ be a monotone net of selfadjoints in $M$ and let $x=\sup x_j\in M$. As ...

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I'm assuming that by $\sigma(\Sigma P)\leq1$ you mean that $\|\Sigma P\|\leq1$. Suppose that $\|P+Q\|\leq1$. So $0\leq P+Q\leq 1$. Then $(P+Q)^2\leq P+Q$ (just conjugate with $(P+Q)^{1/2}$). That is, $$P+Q+QP+PQ\leq P+Q,$$ or $QP+PQ\leq0$. If we conjugate this inequality with $Q$, we get $QPQ+QPQ\leq0$. But $QPQ\geq0$, so $QPQ=0$. Then $$... 3 The assertion P\leq Q means Q-P\geq0. Then$$ 0\leq P(Q-P)P=PQP-P\leq P^2-P=P-P=0. $$So P=PQP. Now we can write this equality as$$0=P-PQP=P(I-Q)P=[(I-Q)P]^*[(I-Q)P],$$so (I-Q)P=0, i.e. P=QP. Taking adjoints, P=PQ. The converse also holds: if P=PQ=QP , then$$Q-P=Q^2-QPQ=Q (I-P)Q\geq0. $$Edit: for the equivalence Q-P\geq0 iff Q-P is a ... 3 The key inequality is |\tau(x)|\leq\tau(|x|). I cannot really follow what Jesse is doing in his first inequality, but all we need to do it take the triangle inequality to get a sum of terms$$ |(\tau\otimes\tau)((p_k\otimes1)(v^*\otimes w^*)U(p_k\otimes 1))|. Then, with x=(p_k\otimes1)(v^*\otimes w^*)U(p_k\otimes 1), \begin{align} ... 2 Suppose that A_1=\{a\in A: \|a\|<1\} and B=\{ b\in A: \|be\|<1\}. Obviously Be\subset A_1, and S_e(Be)=T_e(Be). Hence S_e(Be)\subset T_e(A_1), and consequently \overline{S_e(Be)}^w\subset \overline{T_e(A_1)}^w. But \overline{T_e(A_1)}^w is weakly compact from assumption, and so \overline{S_e(Be)}^w is weakly compact. But ... 2 I will write f_k=r_1+\cdots+r_k, because otherwise we get M_{k+1}(\mathbb C). The projections r_j are each equivalent to 1_A, so there exist partial isometries e_{1,j},\ldots,e_{1,k} with e_{1,j}^*e_{1,j}^\vphantom{*}=r_j,\ \ \ \ e_{1,j}^\vphantom{*}e_{1,j}^*=r_1. $$Now define$$ e_{k,j}^\vphantom{*}=e_{1,k}^*e_{1,j}^\vphantom{*}. $$This is ... 2 What you need to show is that \phi  is one-to-one, and that it is a *-homomorphism. That lets you define the norm on M_n (A)  and the two properties you want are inherited automatically from the codomain. Then you need to show that M_n (A)  is closed. This is done by showing that norm convergence is equivalent to entrywise norm convergence, and as ... 2 First write$$ \lambda I-P = \lambda(I-P+P)-P = \lambda(I-P)+(\lambda-1)P $$If P^{2}=P then (I-P)^{2}=(I-P) and P(I-P)=(I-P)P=0, and the inverse of the above can be spotted for \lambda\in\{0,1\}:$$ (\lambda I-P)^{-1} = \frac{1}{\lambda}(I-P)+\frac{1}{\lambda-1}P. $$Therefore \sigma(P)\subseteq \{0,1\}. And 0\notin\sigma(P) \iff ... 1 The category of C^*-algebras is complete like Martin Brandenburg explained. However, given an inverse system of C^*-algebras, it is often more convenient to consider its limit in the category of all topological *-algebras. The limit of a diagram (A_i,\lVert - \rVert)_{i \in I} of C^*-algebras is then the topological *-subalgebra of the ... 1 You don't need the square, which gives you the additional step of justifying that (1-A)^2\leq 1-A. You can simply do, since 0\leq A\leq 1 (because P\leq A\leq Q),$$ 0\leq P(1-A)P=P-PAP=0, $$so$$ 0=P(1-A)P=[(1-A)^{1/2}P](1-A)^{1/2}P, $$from where (1-A)^{1/2}P=0, and then (1-A)P=0. 1 Perhaps an easier approach is as follows: Let R be any unital commutative ring, then there is a one-to-one correspondence between ideals in R and ideals in M_2(R). In fact, if J \subset M_2(R) is an ideal, then$$ I = \{a \in R : \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \in J\} $$is the corresponding ideal in R such that J = M_2(I). ... 1 Here is an argument for the last part. Assuming we already know that I=\overline{\bigcup_{B}B\cap I}: If I,J are ideals in A and IJ=0, then for any B\in S we have 0=B\cap IJ=(B\cap I)(B\cap J). As B is prime, B\cap I=0 or B\cap J=0. Suppose B_1\cap I\ne0, and B_2\cap J\ne0, with B_1\subset B_2; then B_2\cap I\supseteq B_1\cap ... 1 Our main problem is that, in principle, we cannot just "map the question" to the elements of S by taking intersections. To solve this, we can use an argument similar to the one on the proof of Theorem 6.2.6 of the same book (and I will even use the same notation). Given an ideal I of A, let J=\overline{\left(\bigcup_{B\in S}B\cap I\right)}. Then J ... 1 This doesn't seem to be true. Take e.g. the C^* algebra \mathcal{A} = \mathbb{C}, the complex numbers. Then the element -1 satisfies$$(-1)(-1)^*(-1) = -1 but its spectrum (when viewed as an operator) is $\{-1\}$.

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