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Question:

if $ A_n \to A $ and $ B_n \to B $, show that $ A_n \cup B_n \to A \cup B$ and $ A_n \cap B_n \to A \cap B$

My solution way is the following;

$$ \lim_{n\to \infty} A_n = A $$ and $$ \lim_{n\to \infty} B_n = B $$

$(\liminf A_n)\cup (\liminf B_n) \subseteq \lim inf(A_n \cup B_n)\subseteq\lim sup(A_n \cup B_n)$

$= (\limsup A_n) \cup (\limsup B_n) $

I dont know whether this point correct or not.

$\liminf A_n = \limsup A_n =A$ and $\lim inf B_n = \lim sup B_n =B$

Also how can I prove the following statement

$ \liminf(A_n \cup B_n) \supset \liminf A_n \cup \liminf B_n$

All I can is that! I think that this is not exactly a proper solution. Thus, please show me a sound solution. thank you for helping.

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  • $\begingroup$ Could you please define $A_n$ and $B_n$, are they sequences of sets? $\endgroup$ Feb 28, 2015 at 22:38
  • $\begingroup$ yes they are sequences of sets. @MichaelBurr $\endgroup$
    – 1190
    Feb 28, 2015 at 22:41

1 Answer 1

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Your first part looks good to me. Recall that the elements of $\liminf A_n$ are in all but finitely many $A_n$'s. The elements of $\limsup A_n$ are in infinitely many $A_n$'s. Note that $\liminf A_n\subseteq \limsup A_n$ is always true. When $\lim A_n=A$, then $\limsup A_n=\liminf A_n=A$.

If $x\in(\liminf A_n)\cup(\liminf B_n)$, then it is in all but finitely many of $A_n$ or $B_n$. Therefore, it is in all but finitely many $A_n\cup B_n$ as these are bigger sets. This gives your first line.

You are almost done with the proof at this point. Suppose that $x\in\limsup (A_n\cup B_n)$. This means that $x$ is in infinitely many $A_n\cup B_n$'s. Now, since there are two sets here, $x$ is either in infinitely many $A_n$'s or infinitely many $B_n$'s. Therefore, $x$ is in one of the $\limsup$'s. Hence, $\limsup (A_n\cup B_n)\subseteq (\limsup A_n)\cup(\limsup B_n)$.

The other direction is similar.

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