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Not a duplicate of

$\cap_{A \in \mathcal{F}}(B \cup A) \subseteq B \cup (\cap \mathcal{F})$

This is exercise $3.5.16.b$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $\mathcal F$ is a nonempty family of sets and $B$ is a set. Prove that $B\cup(\bigcap \mathcal F)=\bigcap_{A\in \mathcal F}(B\cup A)$.

Here is my proof:

$(\rightarrow)$ Let $x$ be an arbitrary element of $B\cup(\bigcap\mathcal F)$. Let $A$ be an arbitrary element of $\mathcal F$. Now we consider two different cases.

Case $1.$ Suppose $x\in B$ and so $x\in B\cup A$.

Case $2.$ Suppose $x\in\bigcap\mathcal F$. From $x\in\bigcap\mathcal F$ and $A\in \mathcal F$, $x\in A$ and so $x\in B\cup A$.

Since the above cases are exhaustive, $x\in B\cup A$. Thus if $A\in\mathcal F$ then $x\in B\cup A$. Since $A$ is arbitrary, $\forall A(A\in\mathcal F\rightarrow x\in B\cup A)$ and so $x\in\bigcap_{A\in\mathcal F}(B\cup A)$. Therefore if $x\in B\cup(\bigcap\mathcal F)$ then $x\in\bigcap_{A\in\mathcal F}(B\cup A)$. Since $x$ is arbitrary, $\forall x\Bigr(x\in B\cup(\bigcap\mathcal F)\rightarrow x\in\bigcap_{A\in\mathcal F}(B\cup A)\Bigr)$ and so $B\cup(\bigcap \mathcal F)\subseteq\bigcap_{A\in \mathcal F}(B\cup A)$.

$(\leftarrow)$ Let $x$ be an arbitrary element of $\bigcap_{A\in\mathcal F}(B\cup A)$. We consider two different cases.

Case $1.$ Suppose $x\in\bigcap\mathcal F$. Therefore $x\in B\cup(\bigcap\mathcal F)$.

Case $2.$ Suppose $x\notin \bigcap\mathcal F$. So we can choose some $A_0$ such that $A_0\in\mathcal F$ and $x\notin A_0$. From $x\in\bigcap_{A\in\mathcal F}(B\cup A)$ and $A_0\in\mathcal F$, $x\in B\cup A_0$. From $x\in B\cup A_0$ and $x\notin A_0$, $x\in B$. Therefore $x\in B\cup(\bigcap\mathcal F)$.

Since the above cases are exhaustive, $x\in B\cup(\bigcap\mathcal F)$. Therefore if $x\in\bigcap_{A\in\mathcal F}(B\cup A)$ then $x\in B\cup(\bigcap\mathcal F)$. Since $x$ is arbitrary, $\forall x\Bigr(x\in\bigcap_{A\in\mathcal F}(B\cup A)\rightarrow x\in B\cup(\bigcap\mathcal F)\Bigr)$ and so $\bigcap_{A\in \mathcal F}(B\cup A)\subseteq B\cup(\bigcap \mathcal F)$.

Ergo $B\cup(\bigcap \mathcal F)=\bigcap_{A\in \mathcal F}(B\cup A)$. $Q.E.D.$

Is my proof valid$?$

Thanks for your attention.

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    $\begingroup$ @amWhy I disagree. My proof is different. $\endgroup$ – Khashayar Baghizadeh Jul 17 '20 at 19:14
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    $\begingroup$ @amWhy It's a solution-verification type of question. So does it matter if the question title is even the same$?$ I am self studying the material and do not have access to any real person. I do not think that labeling all my effort a duplicate would be fair. $\endgroup$ – Khashayar Baghizadeh Jul 17 '20 at 19:22
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It’s correct, but Case $1$ of the second part is incomplete: given the level of detail that you’re using elsewhere in the proof, you really should justify the unstated assumption that $\bigcap\mathcal{F}\subseteq\bigcap_{A\in\mathcal{F}}(B\cup A)$. I would reorganize the second part altogether (and shorten it!):

Let $x\in\bigcap_{A\in\mathcal{F}}(B\cup A)$ be arbitrary; then $x\in B\cup A$ for each $A\in\mathcal{F}$. If $x\in B$, then certainly $x\in B\cup\bigcap\mathcal{F}$. If $x\notin B$, then $x\in A$ for each $A\in\mathcal{F}$, so $x\in\bigcap\mathcal{F}$, and again $x\in B\cup\bigcap\mathcal{F}$. Thus, $\bigcap_{A\in\mathcal{F}}(B\cup A)\subseteq B\cup\bigcap\mathcal{F}$.

Further explanation as requested: To begin the second part you assume that $x\in\bigcap_{A\in\mathcal{F}}(B\cup A)$, which is fine. You then consider the cases $x\in\bigcap\mathcal{F}$ and $x\notin\bigcap\mathcal{F}$, but it’s not immediately clear why these are relevant. If there is to be a division into cases at this point, one would expect the cases to derive fairly straightforwardly from the assumption that $x\in\bigcap_{A\in\mathcal{F}}(B\cup A)$, just as in the first part your two cases derive naturally from the assumption that $x\in B\cup\bigcap\mathcal{F}$.

That’s why I first drew the immediate conclusion from $x\in\bigcap_{A\in\mathcal{F}}(B\cup A)$ that $x\in B\cup A$ for each $A\in\mathcal{F}$. Now, because we’re dealing with a union, it’s easy to see what the natural cases are: either $x\in B$, or $x\in A$ for each $A\in\mathcal{F}$. And those two cases match up perfectly with the structure of the target set $B\cup\bigcap\mathcal{F}$,

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  • $\begingroup$ Could you please explain the part on the unstated assumption more$?$ I really appreciate your help. $\endgroup$ – Khashayar Baghizadeh Jul 17 '20 at 19:42
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    $\begingroup$ @KhashayarBaghizadeh: I’ve added a bit more explanation; does it help? $\endgroup$ – Brian M. Scott Jul 17 '20 at 19:50
  • $\begingroup$ I'm still a little confused! $x\in \bigcap_{A\in\mathcal F}(B\cup A)$ means $\forall A(A\in\mathcal F\rightarrow x\in B\cup A)$. Is it correct$?$ If it is, then how from $\forall A(A\in\mathcal F\rightarrow x\in B\cup A)$ and no specific $A_0$ are we justified to conclude $x\in B\cup A$ and then split into cases$?$ $\endgroup$ – Khashayar Baghizadeh Jul 17 '20 at 20:01
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    $\begingroup$ @KhashayarBaghizadeh: By definition $x\in\bigcap_{A\in\mathcal{F}}(B\cup A$ means that $x\in B\cup A$ for every $A\in\mathcal{F}$. For each $A\in\mathcal{F}$ we know that $x\in B$ or $x\in A$, and the first alternative, $x\in B$, is the same for every $A\in\mathcal{F}$. Thus, it’s natural to pull it out and see what happens if $x\in B$, and sure enough, that puts $x$ in the target set. Then we ask what happens if $x\notin B$, and that’s the point at which we actually make use of the universal quantifier: we know that $x$ is in every $A\in\mathcal{F}$. $\endgroup$ – Brian M. Scott Jul 17 '20 at 20:12
  • $\begingroup$ I get it now. Thanks for your time. $\endgroup$ – Khashayar Baghizadeh Jul 17 '20 at 20:18

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