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

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.

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

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, $\limsup (A_n\cup B_n)\subseteq (\limsup A_n)\cup(\limsup B_n)$.