How to use the given condition that $I$ and $J$ are non-empty to prove the given set equality (question in Terrence Tao analysis book)?

Question

The full question is:

Suppose that $I$ and $J$ are two sets, and for all $\alpha ∈ I ∪ J$ let $A_{\alpha}$ be a set.
Show that $(\bigcup_{\alpha\in I} A_{\alpha})\cup (\bigcup_{\alpha\in J} A_{\alpha}) = \bigcup_{\alpha\in I\cup J} A_{\alpha}$.
If $I$ and $J$ are non-empty, show that $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha}) = \bigcap_{\alpha\in I\cup J} A_{\alpha}$.

I managed to prove the first equality without any issues, but when it comes to the second equality, my proof never used the fact that $I$ and $J$ must be non-empty.

My proof:

Now, we show that if $I$ and $J$ are non-empty, then $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha}) = \bigcap_{\alpha\in I\cup J} A_{\alpha}$.

1. First, we show that $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})\subseteq \bigcap_{\alpha\in I\cup J} A_{\alpha}$.

Let $x\in (\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})$.
Then $x\in(\bigcap_{\alpha\in I} A_{\alpha})$ and $x\in (\bigcap_{\alpha\in J} A_{\alpha})$.
By definition of $\bigcap$ , $x\in A_{\alpha}$ for all $\alpha\in I$ and $x\in A_{\alpha}$ for all $\alpha\in J$.
This implies that $x\in A_{\alpha}$ for all $\alpha\in I\cup J$.
By definition of $\bigcap$, it follows that $x\in\bigcap_{\alpha\in I\cup J} A_{\alpha}$.

Thus $x\in (\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})\implies x\in\bigcap_{\alpha\in I\cup J} A_{\alpha}$.
Hence $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})\subseteq \bigcap_{\alpha\in I\cup J} A_{\alpha}$.

2. Now, we show that $\bigcap_{\alpha\in I\cup J} A_{\alpha}\subseteq(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})$.

Let $x\in\bigcap_{\alpha\in I\cup J} A_{\alpha}$.
By definition of $\bigcap$, it follows that $x\in A_{\alpha}$ for all $\alpha\in I\cup J$. This implies that for all $\alpha\in I$ and for all $\alpha\in J$, we have $x\in A_{\alpha}$.
Thus by definition of $\bigcap$, it follows that $x\in(\bigcap_{\alpha\in I} A_{\alpha})$ and $x\in(\bigcap_{\alpha\in J} A_{\alpha})$.
Thus, $x\in(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})$.

We have shown that $x\in\bigcap_{\alpha\in I\cup J} A_{\alpha}\implies x\in (\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})$.

Thus $\bigcap_{\alpha\in I\cup J} A_{\alpha}\subseteq(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})$.

3. Since $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})\subseteq \bigcap_{\alpha\in I\cup J} A_{\alpha}$ and $\bigcap_{\alpha\in I\cup J} A_{\alpha}\subseteq(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha})$, it follows that $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha}) = \bigcap_{\alpha\in I\cup J} A_{\alpha}$. $\square$

My question: Where in the proof should I have used the fact that $I$ and $J$ are non empty sets? Also, is this fact strictly necessary? The statement seems vacuously true if the sets were empty (although I am not sure of this)

Answer 1

When we want Union , we take some element $\alpha$ from $I$ to "generate" the $A_\alpha$ which will also occur when we take the same element $\alpha$ from $I \cup J$ to "generate" the same $A_\alpha$.
Hence we can account for all elements in "Union of Individual Unions" $\bigcup_{\alpha \in I} {A_\alpha}$ , $\bigcup_{\alpha \in J}{A_\alpha}$ along with "Combined Union" $\bigcup_{\alpha \in I \cup J}{A_\alpha}$ to get the first Equality.

When we want Intersection , let $I = J = \phi$ , then the Equality will work out to $\phi \cap \phi = \phi$ , no issues.

The issues will occur when one is empty while the other is not empty.
Let $J = \phi$ , then there is no $\alpha \in J$ , hence no $A_\alpha$ there , hence $\bigcap_{\alpha \in J}{A_\alpha} \equiv \phi$
Hence "Intersection of Individual Intersections" is $\phi$

Checking the "Combined Intersection" , we see that is some $\alpha$ in $I \cup J \equiv I$ , hence there is some $A_\alpha$ , hence $\bigcap_{\alpha \in I \cup J}{A_\alpha}$ is not $\phi$ , generally.

The Equality is lost.

Where you went slightly wrong ?
When you assumed there is some element $x$ : There is no such $x$ in the Case I listed here : The "Intersection of Individual Intersections" has no elements. The "Combined Intersection" will have some such $x$ , in general.

( Comments are useful : user "Ethan Bolker" has already hinted at that , while user "Mauro ALLEGRANZA" has highlighted some issues too )

Answer 2

Recall Tao's definition of the intersection of a family of sets:

We can similarly form intersections of families of sets, as long as the index set is non-empty. More specifically, given any non-empty set $I$, and given an assignment of a set $A_α$ to each $α ∈ I$, we can define the intersection $\bigcap_{α∈I} A_α$ by first choosing some element $β$ of $I$ (which we can do since $I$ is non-empty), and setting $$\bigcap_{α∈I} A_α := \{x ∈ A_β : x ∈ A_α \text{ for all }α ∈ I\}, \tag{3.3}$$ which is a set by the axiom of specification. This definition may look like it depends on the choice of $β$, but it does not (Exercise 3.4.9). Observe that for any object $y$, $$y ∈ \bigcap_{α∈I} A_α \Longleftrightarrow (y ∈ A_α \text{ for all } α ∈ I) \tag{3.4}$$ (compare with (3.2)).

Why does Tao not define the intersection of the empty family indexed by $I = \emptyset$?

Well, which set should $E = \bigcap_{\alpha \in \emptyset}A_\alpha$ be? If you look at $(3.4)$ it should have the property that an object $y$ belongs to $E$ iff $y \in A_\alpha$ for all $\alpha \in \emptyset$. Since there are no such $\alpha$, this condition is vacuously satisfied, thus all objects should belong to $E$. Does there exist a set containing all objects? No.

But perhaps we can make a choice for $E$, for example$E = \emptyset$? This would violate $(3.4)$, but so be it. In that case we would definitely also violate the rule $(\bigcap_{\alpha\in I} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha}) = \bigcap_{\alpha\in I\cup J} A_{\alpha}$. To see this let $I = \emptyset$, $J = \{0\}$ and $A_0$ be any set disjoint from $E$. Then $$(\bigcap_{\alpha\in \emptyset} A_{\alpha})\cap (\bigcap_{\alpha\in J} A_{\alpha}) = E \cap A_0 = \emptyset \ne A_0 = \bigcap_{\alpha\in J} A_{\alpha}. $$

Thus it has good reasons that Tao does not define the "empty intersection".

If one works with subsets $A_\alpha$ of a given set $S$, then one can modify Tao's definition to $$\bigcap_{\alpha\in I} A_{\alpha} = \{x \in S \mid x \in A_\alpha \text{ fior all } \alpha \in I\} \subset S.$$ With this definition $\bigcap_{\alpha\in \emptyset} A_{\alpha} = S$.

Perhaps one should not use the notation $\bigcap_{\alpha\in I} A_{\alpha}$ in that case (as it not the same as Taos's intersection), but something like $\bigcap^S_{\alpha\in I} A_{\alpha}$. Anyway, it is easy to see that for $I \ne \emptyset$ on has $\bigcap^S_{\alpha\in I} A_{\alpha} = \bigcap_{\alpha\in I} A_{\alpha}$.

See also Empty intersection and empty union.