# Prove that $B\cup(\bigcap \mathcal F)=\bigcap_{A\in \mathcal F}(B\cup A)$.

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$$?$$

• @amWhy I disagree. My proof is different. Jul 17, 2020 at 19:14
• @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. Jul 17, 2020 at 19:22

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 $$B\cup\bigcap\mathcal{F}$$,

• Could you please explain the part on the unstated assumption more$?$ I really appreciate your help. Jul 17, 2020 at 19:42
• @KhashayarBaghizadeh: I’ve added a bit more explanation; does it help? Jul 17, 2020 at 19:50
• 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$?$ Jul 17, 2020 at 20:01
• @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 $A\in\mathcal{F}$. Jul 17, 2020 at 20:12
• I get it now. Thanks for your time. Jul 17, 2020 at 20:18