$A_i \subset$ ring $R$ st $\Delta A_i \equiv \{ b-c: b,c \in A_i\} = $ an ideal of $R$, then $\bigcap_i (\Delta A_i) = \Delta (\bigcap_i A_i)$.

Question

Let $A$ be any subset of a ring $R$. $A$ is called a delta generating set of an ideal if $\Delta A = \{b - c: b,c \in A\}$ forms an ideal of $R$.

Let $A_i$ be any collection of delta generating sets in $R$, then it's easy to see that $\Delta (\bigcap_i A_i) \subset \bigcap_i (\Delta A_i)$, like so: $b - c \in $ LHS, with $b,c \in \bigcap_i A_i \implies b-c \in \Delta A_i \ \forall i \implies b-c \in RHS$.

I want to prove the reverse inclusion by induction and taking limits of sets somehow.

Consider $B = \bigcap\limits_{i=1}^{n} (\Delta A_i)$ and $C = \Delta (\bigcap\limits_{i=1}^{n} A_i)$. Clearly, $B \subset C$ for $n = 1$. Let that be the base case for induction. Now assume that $B \subset C$ for $1, \dots, n-1$. And introduce another set $A_i$. Then we have: $\Delta A_n \cap \bigcap\limits_{i=1}^{n-1}(\Delta A_i) \subset \Delta A_n \cap \Delta (\bigcap\limits_{i=1}^{n-1} A_i).$ The RHS is a subset of $\Delta (\bigcap\limits_{i=1}^n A_i)$ since if $b-c \in$ RHS with $b, c \in A_n $ and $b,c \in A_i, i =1, \dots, n-1$, then clearly $b-c \in$ the $ \Delta$ of the intersection of all the sets up to $n$.

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Thus we have for finite collections of sets $A_i \subset R$, that $\Delta (\bigcap\limits_{i=1}^n A_i) = \bigcap\limits_{i=1}^n (\Delta A_i)$. In particular, if the sets $A_i$ are delta generating sets of ideals then $\Delta (\bigcap\limits_{i=1}^n A_i)$ is an ideal.

So is it true for any countable collection of delta generating sets of ideals, by taking limits some how?

Answer 1

Your induction argument doesn't work. If you have $x\in\Delta A_n \cap \Delta (\bigcap\limits_{i=1}^{n-1} A_i)$, all you know is that you can write $x=b-c$ with $b,c\in A_n$, and you can also write $x=d-e$ with $d,e\in \bigcap\limits_{i=1}^{n-1} A_i$. You can't say that $b=d$ and $c=e$.

In fact, the statement is not true for $n=2$. For instance, if $A_1=\{x\}$ and $A_2=\{y\}$ for two distinct elements $x,y\in R$, then $\Delta A_1=\Delta A_2=\{0\}$ is an ideal, but $\Delta(A_1\cap A_2)=\emptyset\neq \Delta A_1\cap \Delta A_2$ (and $\Delta(A_1\cap A_2)$ is not even an ideal).