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.

  • 1
    $\begingroup$ @amWhy I disagree. My proof is different. $\endgroup$ Jul 17, 2020 at 19:14
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 17, 2020 at 19:22

1 Answer 1


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

  • $\begingroup$ Could you please explain the part on the unstated assumption more$?$ I really appreciate your help. $\endgroup$ Jul 17, 2020 at 19:42
  • 1
    $\begingroup$ @KhashayarBaghizadeh: I’ve added a bit more explanation; does it help? $\endgroup$ Jul 17, 2020 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$?$ $\endgroup$ Jul 17, 2020 at 20:01
  • 1
    $\begingroup$ @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 $A\in\mathcal{F}$. $\endgroup$ Jul 17, 2020 at 20:12
  • $\begingroup$ I get it now. Thanks for your time. $\endgroup$ Jul 17, 2020 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.