Is my proof valid? Basic proof for set theory, also is the argument valid if the goal is the other way around?

Question

Suppose $\mathcal{F}$ and $\mathcal{G}$ are nonempty families of sets. Prove that if $\mathcal{F} \subseteq \mathcal{G}$ then $ \bigcap \mathcal{G} \subseteq \bigcap \mathcal{F}$

Proof. Let $x$ and $A$ be arbitrary, suppose $x \in \bigcap \mathcal{G} $ and $A \in \mathcal{F}$ then since $\mathcal{F} \subseteq \mathcal{G}$ it follows that $A \in \mathcal{G}$, so $x \in A$, since $A$ is arbitrary and $A \in \mathcal{F}$ then $ x \in \bigcap \mathcal{F}$, and since $x$ is arbitrary then $ \bigcap \mathcal{G} \subseteq \bigcap \mathcal{F}$

$\square$

Is my proof correct ? Also wouldn't the same argument hold if the goal was to prove $ \bigcap \mathcal{F} \subseteq \bigcap \mathcal{G}$? given again that $\mathcal{F} \subseteq \mathcal{G}$

Something like this:

"Proof". Let $x$ and $A$ be arbitrary, suppose $x \in \bigcap \mathcal{F} $ and $A \in \mathcal{F}$ then since $\mathcal{F} \subseteq \mathcal{G}$ it follows that $A \in \mathcal{G}$, and since $x \in \bigcap \mathcal{F} $, $x \in A$. $A$ is arbitrary and $A \in \mathcal{G}$ then $ x \in \bigcap \mathcal{G}$, and since $x$ is arbitrary then $ \bigcap \mathcal{F} \subseteq \bigcap \mathcal{G}$

$\square$

Many thanks for taking the time to read my question, kind regards.

Answer 1

It seems to me that your confusion comes from your use of the word "arbitrary".

When assuming some property about an arbitrary variable, you can discard the assumption at the end of the proof to say something universal about all objects satisfying the property. For example, when assuming $x\in X$ is arbitrary, then at the end of the proof we can discard this assumption and state something about all elements of $X$.

In the second proof, you start with an arbitrary $A\in \mathcal F$, but later on in the proof, you strengthen this assumption to "$A$ is arbitrary and $A\in \mathcal G$". This isn't true for arbitrary $A\in \mathcal G$, of course, since $\mathcal G$ may contain elements that are not in $\mathcal F$. Since you've strengthened your assumption, in effect you have only proved that $\bigcap \mathcal F\subseteq\bigcap \mathcal G$ in the stronger case where $\mathcal G=\mathcal F$.

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To be precise, this problem with the word arbitrary already happens in the first sentence of the proofs: you start by stating that $A$ is arbitrary in the first sentence, and afterwards you suppose that $A\in\mathcal F$. To take it literal, you start with an arbitrary set $A$, and then you assume more properties of this arbitrary set $A$. Not every arbitrary set $A$ is going to be a member of $\mathcal F$, of course. You want to take an arbitrary [set with the property of being in $\mathcal F$], and not an [arbitrary set] and assume it has the property of being in $\mathcal F$.

To remedy this, I suggest you take care in your proofs to only use the word arbitrary in the same sentence where you define the properties of your variable, and to make very certain that you never start assuming extra properties of the variable after it has been defined. As an example, I would rewrite your first proof as follows:

Suppose $x \in \bigcap \mathcal{G}$ and $A \in \mathcal{F}$ are arbitrary, then since $\mathcal{F} \subseteq \mathcal{G}$ it follows that $A \in \mathcal{G}$, so $x \in A$. Since $A$ is an arbitrary element of $\mathcal F$, then $ x \in \bigcap \mathcal{F}$, and since $x$ is an arbitrary element of $\bigcap\mathcal G$, then $ \bigcap \mathcal{G} \subseteq \bigcap \mathcal{F}$

Another remedy, would be to use the word "any" instead of "arbitrary".

Take any $x\in\bigcap \mathcal{G}$ and any $A\in\mathcal{F}$, then since $\mathcal{F} \subseteq \mathcal{G}$ it follows that $A \in \mathcal{G}$, so $x \in A$. Since $x\in A$ for any $A\in\mathcal F$, then $x \in \bigcap \mathcal{F}$. Since $x\in\bigcap \mathcal F$ for any $x\in\bigcap\mathcal G$, then $ \bigcap \mathcal{G} \subseteq \bigcap \mathcal{F}$.