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In Murphy's book there is the following remark:

Let $I$ be a closed ideal in a $C^*$-algebra $A$ and $J$ a closed ideal in $I$. Then $J$ is also a closed ideal in $A$.

In the book he uses that closed ideals are $C^*$-sub algebras (uses approximate units) and so any positive $b\in J$ has a root $b^{1/2}$ in $J$ and there is an approximate unit $u_\lambda\in I$. It follows for any $a\in A$ that $au_\lambda b\in J$, since $au_\lambda b^{1/2}\in I$ by virtue of $I$ being an ideal in $A$ and then $au_\lambda b^{1/2}b^{1/2}\in J$ from $J$ being an ideal in $I$.

Since $J$ is closed it follows that $\lim_\lambda au_\lambda b = ab$ lies in $J$. Since positive elements span the algebra it follows $J$ is a left ideal, since $J$ is also self adjoint its a right ideal too.

There are two parts of the proof that seem pointless to me:

  1. Why talk about roots? $a u_\lambda$ already lies in $I$ (no need to consider $au_\lambda b^{1/2}$!) so $au_\lambda b$ lies $J$. Since $u_\lambda$ is an approximate unit in $I$ and $J\subset I$ it follows $\lim_\lambda u_\lambda b = b$ and we could be done already.
  2. If you want to use roots then you don't need the approximate unit, if $b=b^{1/2}b^{1/2}$ then $ab=ab^{1/2}b^{1/2}$ where $ab^{1/2}\in I$ because $I$ is an ideal in $A$, it follows then $ab\in J$.

Am I missing something or is the proof actually two proofs mashed together?

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  • $\begingroup$ What number result is this? $\endgroup$ – Aweygan Feb 12 '17 at 16:59
  • $\begingroup$ It is remark 3.1.2. (page 80). $\endgroup$ – s.harp Feb 12 '17 at 17:02
  • $\begingroup$ I think your approach ignoring square roots isn't valid (unless I'm missing something). Sure $au_\lambda\in I$ for each $\lambda$ and $\lim_\lambda b_\lambda=b$, but how do you combine these to get $ab\in J$? Nevertheless, I think your second approach is valid. $\endgroup$ – Aweygan Feb 12 '17 at 17:37
  • $\begingroup$ @Aweygan $au_\lambda\in I$, so $au_\lambda b\in J$ since $J$ is an ideal in in $I$. $\endgroup$ – s.harp Feb 12 '17 at 17:45
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I think you are right. The argument that if $b\in J^+$ then $ab^{1/2}b^{1/2}\in IJ=J$ works fine.

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  • $\begingroup$ Is the other argument also correct? Because then this proof does actually contain two proofs, and one could ascribe a pedagogical motivation to doing it like this. $\endgroup$ – s.harp Feb 13 '17 at 14:37
  • $\begingroup$ Yes, the other argument looks ok to me. Maybe Murphy was thinking about avoiding the common mistake of assuming that $au_\lambda\to a$, but you deal with that fine. $\endgroup$ – Martin Argerami Feb 13 '17 at 15:41

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