Consider the following fragments from Murphy's book '$C^*$-algebras and operator theory'


I'm trying to understand why $B \cap I = BIB$.


The inclusion $BIB \subseteq B\cap I$ is trivial since $B$ is hereditary and $I$ is an ideal. To show the other inclusion, it suffices to show that $(B\cap I)^+ \subseteq BIB$ since the positive elements of the $C^*$-algebra $B \cap I$ (this is a $C^*$-subalgebra because $B \cap I$ is a closed ideal of $B$) linearly span $B\cap I$.

Fix $a \in B \cap I $. Then $a^{1/2} \in B \cap I$.

Let $(u_\lambda)$ be an approximate unit for $B$. Then $$a = \lim_\lambda u_\lambda a = \lim_\lambda {u_\lambda} a^{1/2}a^{1/2} \in BIB$$

Is this correct?


Your argument is fine, but I don't think that the inclusion $BIB\subset B\cap I$ is "trivial". It is even a bit less "trivial" as the other one.

It is obvious that $BIB\subset I$. But you need to show that $ayb\in B$ when $a,b\in B$ and $y\in I$. This requires an approximate unit (as Murphy says): given an approximate unit $\{u_\lambda\}$ in $B$, we can write $y=\sum_{j=1}^4 c_j y_j$ with $y_j\geq0$ for all $j$. Since $0\leq u_\lambda y_j u_\lambda\leq \|y_j\|\,u_\lambda^2\in B$, we get $u_\lambda y_j u_\lambda\in B$ by means of $B$ being hereditary. Thus $u_\lambda y u_\lambda\in B$ for all $\lambda$. Then $$ ayb=\lim_\lambda a(u_\lambda y u_\lambda)b\in B $$ as for each $\lambda$ the three elements in the product are in $B$. As $BIB$ is the closed linear span of elements $ayb$ as above, we get $BIB\subset B$.

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  • $\begingroup$ Yes, you are right but Murphy proved that a $C^*$-subalgebra is hereditary iff $bab'\in B$ for all $b,b'\in B, a\in A$ already. But I should have said I used that result. If I assume that result is it ok? $\endgroup$ – user745578 Jul 29 at 6:45
  • $\begingroup$ Yes, with that and that $I\cap B$ is a closed algebra it is ok. $\endgroup$ – Martin Argerami Jul 29 at 12:53

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