Assume that $B$ is a $C^{\*}$ subalgebra of $A$. We say $B$ is totally non hereditary subalgebra of $A$ if not only $B$ is not a hereditary subalgebra but also it is not isomorphic to any hereditary subalgebra of $A$.


Example: For every $C^{\*}$ algebra $A$, $c(A)$ is a totally non hereditary subalgebra of $\ell^{\infty}(A)$. Here $c(A)$ and $\ell^{\infty}(A)$ are the algebra of all convergent and bounded sequence with $A$-elements, respectively.


There are some $C^{\*}$ algebras which does not have such subalgebras. Example $\mathbb{C}^{n}$



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> To what extent all $C^{\*}$ algebras without totally nonhereditary subalgebras are classified? What are some more examples(other than $\mathbb{C}^{n}$)?
>  **In the topological language, somehow, we are interested in some non trvial example of compact Hausdorff space $X$ such that each quotient of $X$ is homeomorphic to an open subset of $X$**.
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> What is a totally nonhereditary subalgebra of $B(H)$ where $H$ is an infinite dimensional Hilbert space? What is an example of a simple $C^{\*}$ algebra which contains a totally non hereditary sub $C^{\*}$ algebra? Is there any reference about this concept?
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Totally non hereditary $C^{*}$-subalgebras
So you want to consider a C\*-algebra $A$ with the following property: Every sub-C\*-algebra of $A$ is isomorphic to a hereditary sub-C\*-algebra of $A$.


We can distinguish two cases:


1. If $A$ is finite-dimensional, then $A$ has to be commutative. Indeed, assume $A\cong M\_{k\_1}\oplus\ldots\oplus M\_{k\_l}\oplus\mathbb{C}^n$ for $k\_j\geq 2$ (and $j\geq 1$). Then $A$ has a commutative sub-C\*-algebra of the form $\mathbb{C}^{k\_1}\oplus\ldots\oplus \mathbb{C}^{k\_l}\oplus\mathbb{C}^n$, but a commutative hereditary sub-C\*-algebra of $A$ has at most $l+n$ points in its spectrum.
2. If $A$ is infinite-dimensional, then it contains commutative sub-C\*-algebras $C(X)$ where $X$ is any Peano continuum. Such a C\*-algebra has to have `huge' primitive ideal space.


I don't think this property has been considered in the literature before.


Another remark: Every simple C\*-algebra (not equal to $0$ or $\mathbb{C}$) contains a non-simple sub-C\*-algebra. Every hereditary sub-C\*-algebra of a simple C\*-algebra is simple again. Therefore, every (nontrivial) simple C\*-algebra contains a sub-C\*-algebra with the condition you consider.


Similarly, $B(H)$ contains many sub-C\*-algebras that are not isomorphic to a hereditary sub-C\*-algebras of $B(H)$. This is because the primitive ideal space of $B(H)$ has only two points. Thus, any sub-C\*-algebra $A$ of $B(H)$ such that the primitive ideal space of $A$ has at least three points will do.


