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If $A$ is a C*-algebra, we say that a subset $I\subseteq A$ is hereditary if $$ 0\leq x \leq y \in I \Rightarrow x\in I. $$ It is is well known that closed 2-sided ideals are hereditary.

Would it also be true for arbitrary 2-sided ideals? What about self-adjoint 2-sided ideals?

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    $\begingroup$ There are often big families of non-closed ideals which are hereditary though, of which the best known is the Pedersen ideal. $\endgroup$
    – Douglas Somerset
    Commented Oct 1, 2020 at 22:13

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No. Take $A = C[0,1]$ and let $I$ be the (unclosed) ideal generated by the function $f(t) = t$. This ideal is self-adjoint, but it does not contain the function $g(t) = t\sin^2(\frac{1}{t})$, so it is not hereditary. (Example II.5.2.1 (iii) in Bruce Blackadar's fantastic book Operator Algebras: Theory of C${}^*$-Algebras and von Neumann Algebras.)

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  • $\begingroup$ (My first example, which everyone seemed to like, was actually not correct.) $\endgroup$
    – Nik Weaver
    Commented Sep 28, 2020 at 19:11
  • $\begingroup$ Regarding your 1st example I am puzzled. It seems to me that $g=fh$, there, implied that $h$ is constant equal to one on $[0,1/2]$ which is impossible. Can you give me a hint why this is wrong? $\endgroup$
    – Black
    Commented Sep 28, 2020 at 20:52
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    $\begingroup$ Yeah, okay. In that example I was correct in saying that $g$ is neither a scalar multiple of $f$ nor of the form $fh$ for some $h \in A$ $\ldots$ however, it is a sum of such things. So it does belong to the ideal generated by $f$. $\endgroup$
    – Nik Weaver
    Commented Sep 28, 2020 at 21:10

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