# Verify proof that $\varnothing \in F$ implies $\cap F = \varnothing$

Hi so I want to know if my proof attempt is correct, so here's the question:

Suppose that $F$ is a family of sets. Prove that if $\varnothing\in F$ then $\cap F=\varnothing$.

Proof.

Suppose $\varnothing\in F$ and suppose $\cap F\ne\varnothing$. But this means that $x\in\varnothing$, since $\varnothing\in F$. Since $A$ was an arbitrary element of $F$, $x\in\varnothing$. But this is a contradiction since $\varnothing$ is the empty set. Thus $\cap F=\varnothing$.

Therefore if $\varnothing\in F$ then $\cap F=\varnothing$.

• Yeah...its sounds good – Anupam Oct 11 '13 at 11:35
• Hello and welcome to Maths.SE! A tip about MathJax: you don't need to enclose every single symbol in dollar signs, and this makes the code hard to read -- so please don't. A second tip about MSE: this is a quite serious site, so please use proper grammar, spelling and punctuation to the best of your ability. – Lord_Farin Oct 11 '13 at 11:36
• What is $x$ and what is $A$? – Algebraic Pavel Oct 11 '13 at 11:39
• Ok that's what i was worried about, so i just assumed x and A for the definition of $\cap$$F$ not being empty!!! – pkjag Oct 11 '13 at 11:45
• Actually, you could note that more grenerally $A\in F$ implies $\bigcap F\subseteq A$. Therefore $\emptyset\in F$ implies $\bigcap F\subseteq \emptyset$. – Hagen von Eitzen Oct 11 '13 at 11:46

## 3 Answers

Your proof is valid. However, it seems that the sentence starting with "Since $A$" is superfluous -- also, you haven't specified what $A$ is.

One thing that you could improve would be to expand on the second sentence by letting the reader know what $x$ is. E.g.:

But this means that for some $x$, $x \in \varnothing$, since $\varnothing \in F$.

If the course this is homework for is introductory, it is also good to recall the definition of any concept you're using. So you could add a sentence like:

By definition of $\cap F$, we have $x \in \cap F$ for some $x$ if and only if $x \in A$ for every $A \in F$.

to make it explicit where you draw the conclusion $x \in \varnothing$ from.

The introduction of $x$ is a bit jarring, since it's not very clear what it is supposed to be. I would start the first sentence with "Then there is some $x\in \cap F$, but this means…"

The third sentence is superfluous. You have already concluded that $x\in\varnothing$ from the second sentence, so you do not need to do it again.

The last sentence is questionable. The only purpose it serves is to remind the reader what the premises are, because you just recently stated the conclusion. But the proof is so short that the reader is unlikely to have forgotten the premises. As a matter of personal preference, I would end the first paragraph with "Thus $\cap F=\varnothing$, as desired." This is a gentler way to remind the reader that this is what you said at the beginning that you were working toward. But for a proof of this length, even that much of a conclusion is probably not necessary.

The earlier answers have done a fine job of giving feedback on your proof.

For comparison, here is an alternative proof in a more calculational style. Note how this proof starts at the most complex side, expands the definitions to get to the element/logic level, and then from $(*)$ continues using predicate logic. Then from $(**)$ we work our way back to the set level.

\begin{align} & \cap F = \varnothing \\ \equiv & \qquad \text{"basic property of $\;\varnothing\;$"} \\ & \langle \forall x :: x \not\in \cap F \rangle \\ \equiv & \qquad \text{"definition of $\;\cap\;$"} \\ (*)\quad\phantom\equiv & \langle \forall x :: \lnot \langle \forall A : A \in F : x \in A \rangle \rangle \\ \equiv & \qquad \text{"logic: DeMorgan"} \\ & \langle \forall x :: \langle \exists A : A \in F : x \not\in A \rangle \rangle \\ \Leftarrow & \qquad \text{"logic: strengthen using $\;\exists\forall\Rightarrow\forall\exists\;$} \\ & \qquad \phantom"\text{-- the only thing we can do, as far as I can see,} \\ & \qquad \phantom"\text{since we don't know anything about $\;F\;$"} \\ (**)\quad\phantom\equiv & \langle \exists A : A \in F : \langle \forall x :: x \not\in A \rangle \rangle \\ \equiv & \qquad \text{"basic property of $\;\varnothing\;$"} \\ & \langle \exists A : A \in F : A = \varnothing \rangle \\ \equiv & \qquad \text{"logic: one-point rule"} \\ & \varnothing \in F \\ \end{align}