Descending chain of right-annihilators in a ring I have the following problem wich is marked as trivial in the textbook:
Let $R$ be a non-commutative unitary ring, in wich ascending chains of left-annihilators get stationary. Then descending chains of right-annihilators get stationary.
Since $A \subseteq B$  implies $l(B) \subseteq l(A)$, where $l(A) = \{ x \in R | xA = 0 \}$ is the left-annihilator of $A$, I have that if $r(A_{i})$ is a descending chain of right-annihilators, the chain $l(r(A_{i}))$ is an ascending chain of left annihilators so it get stationary: $l(r(A_{n})) = l(r(A_{n+1}))$ for some $n \in \mathbb{N}$. But I fail at showing that this implies $r(A_{n}) \subseteq r(A_{n+1})$. Does somebody know how to prove this?

The important thing about the maps $\ell()$ and $\mathscr r()$ maps is that they form an antitone Galois connection between the poset of left ideals and the poset of right ideals.
One thing this means is that $\ell\mathscr r\ell=\ell$ and $\mathscr{r}\ell\mathscr{r}=\mathscr r$. In particular, if $A$ is a right annihilator ideal, $\mathscr r\ell(A)=A$.
Suppose $A_0\supseteq A_1\supseteq\cdots$ is a descending chain of right annihilators. Then $\ell(A_0)\subseteq \ell(A_1)\subseteq\cdots$ is an ascending chain of left annihilators, and it must become stationary, say at $n$, so that $\ell(A_n)=\ell(A_{m})$ for every $m\geq n$. 
But applying $\mathscr r$ again we have that $\mathscr r\ell(A_n)=\mathscr r\ell(A_m)$ for every $m\geq n$.    By the property I referenced in the second paragraph, $A_n=A_m$ for all $m\geq n$.
You can easily see that this argument works with "left" and "right" interchanged, and also "ascending" and "descending" interchanged.
