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I am studying this proof and there are a few things I need help understanding.

Let $A$ be a set and $B_i$, for $i \in I$ be a family of sets

Prove

$ A \cup (\cap_{i \in I}B_i)$=$\cap_{i \in I}(A \cup B_i)$

proof: suppose $x \in A$ then $x \in (A \cup B_i)$ for all $i \in I$

Also if $x \in \cap_{i \in I}B_i$, $x \in (A \cup B_i)$ for all $i \in I$

thus $x \in \cap_{i \in I}(A \cup B_i)$.

Can someone please explain to me what's in the set $ \cap_{i \in I}(A \cup B_i)$.

Is it the intersection of $A$ included with the intersection of all the $B_i's$

or is it the intersection of all the $B_i's$ with the union of A

In other words is it the intersection of B included with the intersection of the family

Or is it the intersection of the family added with the set A??

I don't get most why the set A can be incorporated in the parenthesis with the intersection of the family!

It just doesn't seem right to me, I need to know if the notation $\cap_{i \in I}B_i$, the intersection of all these sets is restricted to the family.

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    $\begingroup$ $\cap_{i \in I}(A \cup B_i)$ is the set of all elements in $(A \cup B_i)$ for each $i\in I$ $\endgroup$ – saulspatz Jul 14 at 2:50
  • $\begingroup$ @saulspatz so A is not included in any form of intersection right? Is the notation for the intersection restricted to intersecting the family of sets or is A also in that intersection? $\endgroup$ – user686544 Jul 14 at 2:52
  • $\begingroup$ I don't understand what you are asking. What does "A is not included in any form of intersection" mean? $\endgroup$ – saulspatz Jul 14 at 2:54
  • $\begingroup$ @saulspatz I was wondering if the set $ \cap_{i \in I}(A \cup B_i)$ contains either the elements from the intersection of the family intersected with the elements in the loan set A, or is it the set intersection of the family included with every element of the set A? $\endgroup$ – user686544 Jul 14 at 3:07
  • $\begingroup$ @saulspatz I would assume its the latter $\endgroup$ – user686544 Jul 14 at 3:10
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I don't get most why the set A can be incorporated in the parenthesis with the intersection of the family!

How you state the question: "Let $A$ be a set and $B_i$, for $i\in I$ be a family of sets", it is not clear, why you can make 'sense' out of $A\cup B_i$ or other set operations. I suppose, that $A$ and $B_i$ are subsets of the same set $X$. Else you can not really compare these sets in general, and the results might be trivial.

So you might check that first.

$\bigcap_{i\in I} (A\cup B_i)$ is the intersection of $(A\cup B_i)$ for every $i\in I$.

For example: Take $I=\{1,2,3\}$ And let $A, B_1,B_2, B_3\subseteq X$ with $X=\{1,2,3\}$. And $A=\{1,2,3\}, B_1=\{1\}, B_2=\{1,2\}, B_3=\{1,2,3\}$, then:

$\bigcap_{i=1}^3 (A\cup B_i)=(A\cup B_1)\cap (A\cup B_2)\cap (A\cup B_3)=A$.

The example given is not really insightful, but I just wanted to show, how this intersection works.

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$x\in\bigcap_{i\in I}(A\cup B_i)$ exactly when $\forall i\in I~[x\in(A\cup B_i)]$. That is $\forall i\in I~[x\in A\lor x\in B_i]$.

In words: For all $i$ in the indices $I$, $x$ is in $A$ or $x$ is in $B_i$.

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  • $\begingroup$ So essentially $ \cap_{i \in I}(A \cup B_i)$ is all the elements in the intersection of all the sets in the family added with all the elements of the set A? $\endgroup$ – user686544 Jul 14 at 2:57
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    $\begingroup$ That is what it says, yes. Essentially: $(A\cup B_1)\cap(A\cup B_2)\cap\cdots$ $\endgroup$ – Graham Kemp Jul 14 at 2:58

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