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

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

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.

• Your proofs seems correct and are pretty much the same as your books' proof, with a bit different wording. Apr 15, 2017 at 20:16
• Your proof is much better than the proof in your book. The use of the law of the excluded middle in the book is unnecessary. Please name the book. Apr 15, 2017 at 20:35
• There's a little mistake in the first proof. The union should be an intersection.
– user370967
Apr 15, 2017 at 20:52
• @RobArthan Well, I didn't expect that (maybe it's my translation). The book was written in spanish by my teacher for a first-year course called "fundamentos de matemáticas" (math fundamentals or foundations of mathematics), you can download it from here: Fundamentos de Matemáticas. The last proof shown here is at the end of page 57. Apr 15, 2017 at 20:54
• Thanks for the link. I am sure it was just an oversight on the part of your teacher in that proof. Given $x \in B \cup A_i$, you don't need to know that $x \not\in B$ when you deal with the case when $x \in A_i$. Apr 15, 2017 at 21:03

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.