[proofplan] We use the quotient map rather than constructing approximate factorizations directly. Since $I$ is a closed two-sided ideal, the quotient $A/I$ is a C*-algebra by the [quotient C*-algebra theorem](/theorems/8568). If $0\le a\le b$ and $b\in I$, then the image of $a$ in $A/I$ is both positive and negative, hence zero. Therefore $a$ lies in the kernel of the quotient map, which is exactly $I$. [/proofplan] [step:Pass to the C*-quotient by the closed ideal] Let \begin{align*} q:A&\to A/I \end{align*} denote the quotient map, given by $q(x)=x+I$ for $x\in A$. Since $I\trianglelefteq A$ is a closed two-sided ideal, [citetheorem:8568] implies that $A/I$, equipped with the quotient norm, quotient multiplication, and quotient involution, is a C*-algebra. In particular, $q$ is a surjective $*$-homomorphism and \begin{align*} \ker q=I. \end{align*} [/step] [step:Show that the positive element maps to zero in the quotient] Let $a,b\in A_+$ satisfy $0\le a\le b$ and $b\in I$. The inequality $0\le a\le b$ means that $a\in A_+$ and $b-a\in A_+$. Since $q$ is a $*$-homomorphism, it sends positive elements to positive elements: if $x\in A_+$, then $x=y^*y$ for some $y\in A$, and hence \begin{align*} q(x)=q(y^*y)=q(y)^*q(y)\in (A/I)_+. \end{align*} Applying this to $a$ and $b-a$, we obtain \begin{align*} q(a)\in (A/I)_+ \end{align*} and \begin{align*} q(b-a)\in (A/I)_+. \end{align*} Since $b\in I=\ker q$, we have $q(b)=0$. Therefore \begin{align*} q(b-a)=q(b)-q(a)=-q(a). \end{align*} Thus both $q(a)$ and $-q(a)$ are positive in the C*-algebra $A/I$. [guided] We want to prove $a\in I$, and the quotient map is designed precisely to test membership in $I$: an element belongs to $I$ exactly when its coset in $A/I$ is zero. So we apply \begin{align*} q:A\to A/I \end{align*} to the order inequality $0\le a\le b$. The inequality $0\le a\le b$ has two pieces of content. First, $a$ is positive. Second, $b-a$ is positive. Because $q$ is a $*$-homomorphism, it preserves positivity: whenever $x\in A_+$, there is some $y\in A$ with \begin{align*} x=y^*y, \end{align*} and then \begin{align*} q(x)=q(y^*y)=q(y)^*q(y), \end{align*} which is positive in $A/I$. Applying this to $x=a$ gives \begin{align*} q(a)\in (A/I)_+. \end{align*} Applying it to $x=b-a$ gives \begin{align*} q(b-a)\in (A/I)_+. \end{align*} Now the hypothesis $b\in I$ is used. Since $\ker q=I$, this gives \begin{align*} q(b)=0. \end{align*} By linearity of $q$, \begin{align*} q(b-a)=q(b)-q(a)=-q(a). \end{align*} Hence $-q(a)$ is positive as well. We have reached the decisive order-theoretic situation: the same self-adjoint element $q(a)$ is positive and its negative is also positive. [/guided] [/step] [step:Use properness of the positive cone to conclude membership in the ideal] In any C*-algebra, the positive cone is proper: if an element $z$ satisfies $z\ge 0$ and $-z\ge 0$, then $z=0$. Applying this in the C*-algebra $A/I$ to $z=q(a)$ gives \begin{align*} q(a)=0. \end{align*} Since $\ker q=I$, it follows that $a\in I$. We have shown that for every $a,b\in A_+$ with $0\le a\le b$ and $b\in I$, one has $a\in I$. This is exactly the order-theoretic hereditary condition for $I$ as a C*-subalgebra of $A$. Therefore $I$ is a hereditary C*-subalgebra of $A$. [/step]