# Is every 2-sided ideal in a C*-algebra hereditary?

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?

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

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.)
• 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? – Black Sep 28 at 20:52
• 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 $f$. – Nik Weaver Sep 28 at 21:10