Prove "$\bigcap_{i\in I} (B \cup A_i) \subseteq B \cup (\bigcap_{i\in I} A_i)$" without excluded middle

Question

"$\bigcap_{i\in I} (B \cup A_i) \subseteq B \cup (\bigcap_{i\in I} A_i)$"

Proof. Let $x\in \bigcap_{i\in I} (B \cup A_i)$, then $x\in (B\cup A_i)$, for all $i\in I$. Which implies that $x\in B \vee x\in A_i$, for all $i\in I$. Thus $x\in B \vee x\in (\bigcap_{i\in I} A_i) \Rightarrow x\in B \cup (\bigcap_{i\in I} A_i). \Box$

Another proof I wrote, but longer (I use cases explicitly) is:

Proof. Suppose $x\in \bigcap_{i\in I} (B \cup A_i)$, then $x\in (B\cup A_i)$, for all $i\in I$. Hence $x\in B$ or $x\in A_i$, for all $i\in I$. If $x\in B$, then $x\in B \cup (\bigcap_{i\in I} A_i)$. If $x\in A_i$, for all $i\in I$, then $x\in \bigcap_{i\in I} A_i$. Therefore $x\in B \cup \bigcap_{i\in I} A_i$. Either way, $\bigcap_{i\in I} (B \cup A_i) \subseteq B \cup (\bigcap_{i\in I} A_i). \Box$

Are these proofs correct? Give me suggestions to improve them.

If you're curious, the proof in my book was (more or less) this (the proofs above are my alternatives):

Proof. Let $x\in \bigcap_{i\in I} (B \cup A_i)$, then $x\in (B\cup A_i)$, for all $i\in I.$ There are two possible cases: $x\in B$ or $x\notin B$. If $x\in B$, then $x\in B \cup (\bigcap_{i\in I} A_i)$, since $B \subseteq B \cup (\bigcap_{i\in I} A_i)$. If $x\notin B$, because $x\in (B \cup A_i)$, for all $i\in I$, necessarily $x\in A_i$, for all $i\in I$. Thus $x\in \bigcap_{i\in I} A_i \Rightarrow x\in B \cup (\bigcap_{i\in I} A_i). \Box$

Thank you.

Answer 1

There is a little mistake in your first proof. The $\bigcup$ in the end should be a $\bigcap$, but I'm sure that's what you meant.

For the rest, your proofs are correct, and I prefer the first one out of the three proofs you wrote down as it is not neccesary to consider apart cases here.

Keep up the good work!