Infinite sum of ideals

Solution 1:

The definition is correct. Now it's just a matter of finding the best notation for proving the closure of the set with respect to sums, because the fact that $0$ belongs to the set is obvious and it's also obvious that, if $x$ is in the set and $r\in R$, then $xr$ and $rx$ are in the set.

Instead of using $\beta_i$, it's preferable to write $$ x=\sum_{\beta\in C}x_\beta $$ where $C$ is a finite subset of $B$ and $x_\beta\in I_\beta$. Now, if you have $x$ written in that way and $E$ is a finite subset of $B$ containing $C$, you can also write $$ x=\sum_{\beta\in E}x'_\beta $$ where $$ x'_\beta=\begin{cases} x_\beta & \text{if $\beta\in C$}\\[2ex] 0 & \text{if $\beta\in E\setminus C$} \end{cases} $$ So, if $$ x=\sum_{\beta\in C}x_\beta, \quad y=\sum_{\beta\in D}y_\beta $$ are elements of your set, consider $E=C\cup D$ and rewrite $$ x=\sum_{\beta\in E}x'_\beta, \quad y=\sum_{\beta\in E}y'_\beta $$ as before. Thus $$ x+y=\sum_{\beta\in E}(x'_\beta+y'_\beta) $$ which belongs to the set, because $x'_\beta+y'_\beta\in I_\beta$ for each $\beta\in E$.