Prove that $A_n \cap B_n \rightarrow A \cap B$

Question

If $A_n \rightarrow A$ and $B_n \rightarrow B$ are sequences of sets then is it true that $A_n \cap B_n \rightarrow A \cap B$? How to prove or provide a counterexample?

I had thought the following possible solution, but I am not convinced about it.

I tried to prove the following statements:

(i) $\liminf \left(A_n \cap B_n\right) = A \cap B$;

(ii) $\limsup \left(A_n \cap B_n\right) = A \cap B$;

thus, we can conclude that $\lim (A_n \cap B_n)$, i.e., $A_n \cap B_n \rightarrow A \cap B$

To prove these statements I use the following arguments:

(i) $\liminf \left(A_n \cap B_n\right) = \liminf A_n \cap \liminf B_n = A \cap B$;

(ii) $\limsup \left(A_n \cap B_n\right) \subset \limsup A_n \cap \limsup B_n = A \cap B$;

Therefore, $A_n \cap B_n \rightarrow A \cap B$.

Obs: A similar question was asked here, but the response is not complete. Can anybody help me? Thanks!

Answer 1

By some fundamental union and intersection laws, and the de Morgan laws we have: $$ \liminf (A_k \cap B_k) = \bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty} (A_k \cap B_k) = \bigcup_{n=0}^{\infty} \left(\bigcap_{k=n}^{\infty} A_k \cap \bigcap_{k=n}^{\infty} B_k\right) = \bigcup_{n=0}^{\infty} \bigcap_{k=n}^{\infty} A_k \cap \bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty} B_k \\ = \liminf A_k \cap \liminf B_k = A \cap B . $$

$$ \limsup(A_k \cap B_k) = \bigcap_{n=0}^{\infty} \bigcup_{k=n}^{\infty} (A_k \cap B_k) \subseteq \bigcap_{n=0}^{\infty} \left( \bigcup_{k=n}^{\infty} A_k \cap \bigcup_{k=n}^{\infty} B_k \right) = \bigcap_{n=0}^{\infty} \bigcup_{k=n}^{\infty} A_k \cap \bigcap_{n=0}^{\infty} \bigcup_{k=n}^{\infty} B_k \\ = \limsup A_k \cap \limsup B_k = A \cap B . $$

Thus, $$ A \cap B = \liminf (A_k \cap B_k) \subseteq \limsup (A_k \cap B_k) \subseteq A \cap B, $$ which implies $$ A \cap B = \liminf (A_k \cap B_k) = \limsup (A_k \cap B_k), $$ i.e. $$ A_k \cap B_k \to A \cap B. $$

Answer 2

The answer to your question comes trivially from the following lemma.

Lemma. $D_n\rightarrow D \Leftrightarrow 1_{D_n}\rightarrow 1_D$

Proof. Let us first show $\Rightarrow$.

Assume that $x\in D$. We need to show that $1_{D_n}(x)\rightarrow 1$. Note that $D=\bigcup_k \bigcap_{n\geq k} D_n$ and therefore, $x\in \bigcup_k \bigcap_{n\geq k} D_n$, i.e., $x\in D_n$ eventually. Therefore, $1_{D_n}(x)=1$ eventually and $1_{D_n}(x)\rightarrow 1$ holds.

Now, assume $x\notin D$. We need to show that $1_{D_n}(x)\rightarrow 0$. Note that $D_n^{c}\rightarrow D^{c}$, where $D^c$ is the complement of $D$. From the previous paragraph we conclude that $1_{D_n^c}(x)\rightarrow 1$ and thus, $1_{D_n}(x)\rightarrow 0$.

We now show $\Leftarrow$.

Let $x\in D$. We have that $1_{D_n}(x)\rightarrow 1$. In particular, $x\in D_n$ eventually. In other words, $D\subseteq \bigcup_{k} \bigcap_{n\geq k} D_n$. Now, we will show that $\bigcap_{k} \bigcup_{n\geq k} D_n\subseteq D$. Consider $x\in D_n$ infinitely often. Then, if $x\notin D$ we would have $1_{D_n}(x)=1$ infinitely often while $1_{D}(x)=0$ which contradicts the fact that $1_{D_n}(x)\rightarrow 1_D(x)$ for all $x$. We then conclude that $x\in D$ or in other words,

$\limsup_n D_n:=\bigcap_{k} \bigcup_{n\geq k} D_n\subseteq D \subseteq \bigcup_{k} \bigcap_{n\geq k} D_n=:\liminf_n D_n$.

This implies $D=\liminf_n D_n=\limsup_n D_n$, i.e., $D_n\rightarrow D$.