Show $\left(\bigcup\limits_{i=1}^{\infty} A_{i}\right) \cap B = \bigcup\limits_{i=1}^{\infty} (A_{i}\cap B)$

Question

So far my proof is as follows:

We first show $\left(\bigcup\limits_{i=1}^{\infty} A_{i}\right) \cap B \subset \bigcup\limits_{i=1}^{\infty} (A_{i}\cap B)$. Take $x \in \left(\bigcup\limits_{i=1}^{\infty} A_{i}\right) \cap B$. Then, $x \in \bigcup\limits_{i=1}^{\infty} A_{i}$ and $x \in B$. This means $x \in A_1$ and $B$ or $x \in A_2$ and $B$ or $x \in A_3$ and $B$, etc. So, $x \in \bigcup\limits_{i=1}^{\infty} (A_{i}\cap B)$.

The converse has similar reasoning as above, only in the reverse order. I'm not sure if my proof is thorough enough.

Answer 1

Seems fine, though I will write it more compactly.

$$x \in \left( \bigcup_{i=1}^\infty A_i\right) \cap B$$

$$\iff x \in \left( \bigcup_{i=1}^\infty A_i\right) \land x \in B$$

$$\iff \exists i \in \mathbb{N}, x \in A_i \land x \in B $$

$$\iff \exists i \in \mathbb{N}, x \in A_i \cap B$$

$$\iff x \in \left(\bigcup_{i=1}^\infty A_i \cap B\right)$$

Answer 2

It looks good to me, but I would write it as follows, starting from "Then, $x\in\bigcup_{i=1}^{\infty} A_i$ and $x\in B$":

Since $x\in \bigcup_{i=1}^{\infty} A_i$, we have $x\in A_i$ for some $i\in\mathbb{N}$. Thus $x\in A_i\cap B$, and thus $x\in\bigcup_{i=1}^{\infty}(A_i\cap B)$.

Answer 3

You have the right idea, but it could be written a little better. The key idea is the existence of an index $i$ such that $x \in A_i$. This is a lot more precise then "well, it could be any one of these infinitely many cases...." So a more concise rewriting of this would be:

Let $x \in \left(\bigcup_{i = 1}^{\infty} A_i\right) \cap B$. Then $x \in B$ and there exists an $i^* \in \mathbb{N}$ such that $x \in A_{i^*}$. Then $x \in A_{i^*} \cap B$, and so $x \in \bigcup_{i = 1}^{\infty} (A_i \cap B)$.