Union of ascending ideals is an ideal Could you tell me what I'm doing wrong in proving this proposition?
If $I_1 \subset I_2 \subset ... \subset I_n \subset ...$ is an ascending chain of ideals in $R$, then $I := \bigcup _{n \in \mathbb{N}} I_n$ is also an ideal in $R$.
So we need to prove that the group $(I, +)$ is abelian and if $a \in I$, then $ar, ra \in R$ (but I suppose we could assume that $R$ is commutative).
Here are my attempts: $a \in I \Rightarrow \exists n\in \mathbb{N} : a \in I_n \Rightarrow ar, ra \in I_n \Rightarrow ar, ra \in I$. But I think I don't use the fact that the chain is ascending here. Could you point out my mistake and help me correct it?
Thank you a lot! 

The following is in the book Algebra, by T. Hungerford. 
Let $b \in I_i$ and $c \in I_j$. It follows that $i \leq j$ or $j \leq i$. Say $i \geq j$, then $I_j \subset I_i$. Therefore, $b, c \in I_i$. Since $I_i$ is an ideal, $b-c \in I_i \subset I$. 
Similarly, if $r \in R$ and $b \in I$, $b \in I_i$ for some $i$. Therefore $rb \in I_i \subset I, br \in I_i \subset I$. Hence $I$ is an ideal in $R$. 
The result follows from a theorem previously proved in the book:
A nonempty subset I of a ring R is a left [resp. right] ideal if, and only if, for all $a,b \in I$, $r \in R$:
$a, b \in I \Rightarrow a - b \in I $ and $a \in I, r \in R \Rightarrow ra \in I$ [resp. $ar \in I$]   
