A Proof Involving an Increasing Chain of Ideals in a C$^{*}$-Algebra in Davidson's Book

Question

I am having trouble understanding how the following underlined passage follows:

Background: This is from the proof concerning the correspondence of ideals in AF-algebras with directed hereditary sets of the Bratteli diagrams at the top of page 86 of Davidson's C$^{*}$-algebras book.

Basically: (all ideals are closed two-sided ideals)

$\mathfrak{U}$ is an AF-algebra: $\mathfrak{U}=\overline{\bigcup_{p=1}^{\infty}\mathfrak{U_{p}}}$, where the $\mathfrak{U_{p}}$ are an increasing sequence of finite dimensional C$^{*}$-algebras. $\mathfrak{J_{p}}$ is an ideal in $\mathfrak{U_{p}}$, $\mathfrak{J_{1}}\subset\mathfrak{J_{2}}\subset\cdots$, and $\mathfrak{J}=\overline{\bigcup_{p=1}^{\infty}\mathfrak{J_{p}}}$ is an ideal in $\mathfrak{U}$. We know that $\mathfrak{J_{p}}=\mathfrak{U_{p}}\cap \mathfrak{J_{p+1}}$.

I am having trouble understanding how we can use this to conclude that $\mathfrak{J_{p}}=\mathfrak{U_{p}}\cap\mathfrak{J}$.

For $j\geq p$, we can deduce that $$ \mathfrak{J_{p}}=\mathfrak{U_{p}}\cap \mathfrak{J_{p+1}} =\mathfrak{U}_{p}\cap\mathfrak{U_{p+1}}\cap\mathfrak{J_{p+2}}=\cdots=\mathfrak{U_{p}}\cap\cdots\cap\mathfrak{U_{j-1}}\cap\mathfrak{J_{j}}=\mathfrak{U_{p}}\cap\mathfrak{J_{j}}, $$ from which we can conclude that $$ \mathfrak{U_{p}}\cap\bigcup_{j\geq p}\mathfrak{J_{j}}=\mathfrak{J_{p}}. $$ However, I don't know how to obtain that $$ \mathfrak{U_{p}}\cap\overline{\bigcup_{j\geq p}\mathfrak{J_{j}}}=\mathfrak{J_{p}}, $$ which is what the text claims.

Any help would really be appreciated.

Thank you.

Answer 1

You can use the fact that $U_p\cap J = U_pJ$, the linear span of elements of the form $xy$ where $x\in U_p$ and $y\in J$. You have already shown that $J_p\subset U_p\cap J$, so it suffices to show the reverse inclusion. Hence, it suffices to show that if $x\in U_p, y\in J$, then $xy\in J_p$.

For $\epsilon > 0, \exists z\in J_j$ for $j$ large enough such that $\|y-z\| < \epsilon/\|x\|$, so that $\|xy-xz\| < \epsilon$. By what you have shown, $xz\in U_p\cap J_j \subset J_p$, so since $J_p$ is closed, $xy\in J_p$ as well.