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The Boolean algebra of connected components is equivalent to the projections (the elements with $p^2=p$) in the algebra of functions, with multiplication of functions representing intersection and $(p_1,p_2) \to p_1 + p_2 - p_1p_2$ being the union of sets of components. I think there is a version of algebraic K-theory for topological algebras whose value ...

3

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 ...

2

Let $E=\{1,1/2,1/3,\cdots\}$ and let $\mu_{a}$ be the finite atomic meausre on $[0,1]$ that is supported on $E$ with $\mu\{1/n\}=1/n^{2}$. Let $\mu = \mu_{a}+m$ where $m$ is Lebesgue measure on $[0,1]$. Let $X=L^{2}_{\mu}[0,1]$. Let $\chi$ be the characteristic function of $E$, and define operators $A, B \in \mathcal{L}(X)$ by $$Af = ... 1 Let X be a compact Hausdorff space with more than one point. Then C(X) is a natural, normal uniform algebra on X. This means that the character space of C(X) is exactly X and for each closed set E\subseteq X and each closed set F\subseteq X\setminus E, there exists a function f\in C(X) such that f(y)=0 for all y\in E and f(y)=1 for each ... 1 Yes and you don't even have to take the closure. For instance if A is a non-simple C*-algebra with trivial cetnre that has unique (faithful) trace (for instance A=C^*(G) for a sufficiently non-commutative amenable group such as the group of permutations of integers that move at most finitely many entries), then A\otimes \mathcal{Z}, where \mathcal{Z} ... 1 Kaplansky's density theorem lets you choose a  with \|a\|\leq\|u\|. 1 Since you have to relate the norm with the spectrum, I don't think this has an elementary algebraic proof. Facts needed (from the theory of Banach algebras): For any A\in\mathcal S, \sigma(A)=\{\phi(A):\ \phi\in S(\mathcal A)\}, where S(\mathcal A) is the state space. For any A\in\mathcal S, \|A\|=\text{spr}(A)=\max\{|\lambda|:\ ... 1 It is correct for positive operators, for more information see Gohberg, Krein, Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space page 27 1 Let L denote the matrix$$ L = \pmatrix{u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3} $$My best guess is that whatever the author is getting at has something to do with the fact that$$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_N $$or, if N is odd,$$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + ...

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