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

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.

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    $\begingroup$ Your proofs seems correct and are pretty much the same as your books' proof, with a bit different wording. $\endgroup$
    – AsafHaas
    Apr 15, 2017 at 20:16
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    $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Apr 15, 2017 at 20:35
  • $\begingroup$ There's a little mistake in the first proof. The union should be an intersection. $\endgroup$
    – user370967
    Apr 15, 2017 at 20:52
  • $\begingroup$ @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. $\endgroup$
    – mathman
    Apr 15, 2017 at 20:54
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    $\begingroup$ 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 $x \in A_i$. $\endgroup$
    – Rob Arthan
    Apr 15, 2017 at 21:03

1 Answer 1

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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!

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