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Does convergence of distinct sequences of sets imply anything about the convergence of their unions and intersections. I.e. if $A_n \rightarrow A$ and $B_n \rightarrow B$ does this imply that $A_n \cup B_n \rightarrow A \cup B$ and $A_nB_n \rightarrow AB$?

Gaunnett
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  • To discuss convergence you have to have a metric. I'd assume you're using the metric $d(A,B) = \mu(A \setminus B) + \mu(B \setminus A)$ for a measure $\mu$? Also what is $AB$? The set ${ab : a \in A, b \in B}$? – Anthony Peter Sep 15 '15 at 01:53
  • I'm thinking about set convergence defined as the common value of the $limsup$ and $liminf$ which has the usual intersection/union definitions. $AB$ is just short-hand for intersection. – Gaunnett Sep 15 '15 at 02:05
  • @StudentGosset How do you define $\lim\sup A_n$ of the sequence of sets $A_n$? –  Sep 15 '15 at 03:12
  • For everyone wondering about set limits and convergence - http://math.stackexchange.com/a/107938/21693 – ivancho Sep 15 '15 at 04:45

1 Answers1

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Yes, those convergences hold.

The definition of $A_n \rightarrow A$ is that each element of $A$ belongs to all except finitely many $A_n$ and that the elements of $A$ are the only elements which belong to infinitely many $A_n$.

So let's start with $x \in A \cup B$. It must be either from $A$ or from $B$ - WLOG let's say it's from $A$. Then $x \in A_n$ for all $n$ after some point, so $x \in A_n \cup B_n$ for those $n$, and is therefore in $\liminf A_n \cup B_n$. Conversely, if $x \in A_n \cup B_n$ for infinitely many $n$, then either $x \in A_n$ or $x \in B_n$ for infinitely many $n$, so $x \in A$ or $x \in B$, so $x \in A \cup B$.

$A \cap B$ follows the same way - if $x \in A \cap B$, then $x \in A$ and $x \in B$, so $x \in A_n \forall n>n_1$ and $x \in B_n \forall n>n_2$, and so $x \in A_n \cap B_n \forall n > max(n_1, n_2)$. Conversely, if $x \in A_n \cap B_n$ for infinitely many $n$, then it's in $A_n$ and $B_n$ for infinitely many $n$, so it's in $A$ and $B$, so it's in $A \cap B$

ivancho
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