Empty intersection and empty union

Question

If $A_\alpha$ are subsets of a set $S$ then

$\bigcup_{\alpha \in I}A_\alpha$ = "all $x \in S$ so that $x$ is in at least one $A_\alpha$"

$\bigcap_{\alpha \in I} A_\alpha$ = "all $x \in S$ so that $x$ is in all $A_\alpha$"

It is the convention that $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$ and $\bigcap_{\alpha \in \emptyset} A_\alpha = S$.

But if $x$ is in $\bigcap_{\alpha \in \emptyset} A_\alpha = S$ then $x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha \in \emptyset$. But then $x \in \bigcup_{\alpha \in I}A_\alpha$.

Can someone help me and tell me what is wrong with this? Thank you.

Answer 1

Some texts consider it a convention, but it is in fact a computation!

For the fixed set $S$, we are looking at the set $\mathcal P(S)$, the set of all subsets of $S$. With the operations of intersection and union (of arbitrary many subsets) the set $\mathcal P(S)$ is what is known as a complete lattice (don't worry if you don't know what that means).

First, one can argue intuitively: the more subsets of $S$ you intersect, the smaller the intersection is. Or, the fewer subsets you intersect, the larger the intersection is. So, the intersection of no subsets at all, the least amount of sets you can intersect, should be the largest subset possible. Namely, $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the fewer subsets you take the union of, the smaller the union. So, the union of no subsets at all should be the smallest set possible. Namely, $\bigcup _{i\in \emptyset }A_i=\emptyset$.

Now, to make things more formal, let's define the intersection and union in $\mathcal P(S)$. The definition will be equivalent to the set-theoretic definitions but will only make use of the partial order relation of inclusion. Given a collection $\{A_i\}_{i\in I}$ of subsets of $S$, their intersection is the largest subset of $S$ that is contained in each $A_i$ (notice that this is saying that the intersection is a greatest lower bound). Similarly, the union of the family of subsets is the smallest subset of $S$ that contains each $A_i$ (notice that this says that the union is a least upper bound). Incidentally, this point of view very clearly points to a duality between union and intersection.

So now, the intersection of no subsets is the largest subset of $S$ that is contained in each one of the given subsets. There are no given subsets at all, so (vacuously) any subset $B\subseteq S$ is contained in each of the non-existent $A_i$. The largest of those is $S$, proving that $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the union of no subsets is the smallest subset of S that contains each of the given subsets. No subsets are given, so any subsets $B\subseteq S$ contains each of the non-existent $A_i$. The smallest of these is $\emptyset $, thus proving that $\bigcup_{i\in \emptyset}A_i=\emptyset$.

Answer 2

Please note that the fact that

$$\bigcap_{\alpha \in \emptyset} A_\alpha = S$$

is not a convention - it follows from the definition

$$\bigcap_{\alpha \in I} A_\alpha = \{ x \in S : \text{$x \in A_\alpha$, for all $\alpha \in I$}\}.$$

This is best understood by asking when is it that for a given $x \in S$ we have $x \notin \bigcap_{\alpha \in \emptyset} A_\alpha$. This can only happen if there exists $\alpha \in \emptyset$ such that $x \notin A_{\alpha}$. Since $\emptyset$ has no elements, there cannot be such an $\alpha$.

PS One can avoid introducing a (somewhat arbitrary) index set in an intersection, say. Instead, fix a set $S$, take a subset $\mathfrak{S}$ of $\mathcal{P}(S)$, and define $$ \bigcap \mathfrak{S} = \{ x \in S : \text{$x \in A$ for all $A \in \mathfrak{S}$} \}. $$ Then the argument above becomes $$ \bigcap \emptyset = \{ x \in S : \text{$x \in A$ for all $A \in \emptyset$} \} = \{ x \in S : \ \} = S,$$ since there are no $A$ to consider here.

Answer 3

Some symbolic manipulation can help, to see what is NOT in those sets,

$ x \in \ \bigcap_{\alpha \in I} A_\alpha \ \Longleftrightarrow \forall\alpha\in I,\ x\in A_\alpha \\ x \in \ \bigcap_{\alpha \in I} A_\alpha \ \Longleftrightarrow \forall\alpha, \ \alpha\in I\rightarrow\ x\in A_\alpha \\ x \notin \ \bigcap_{\alpha \in I} A_\alpha \ \Longleftrightarrow \exists\alpha: \ \alpha\in I\ \wedge\ x\notin A_\alpha \\ x \notin \ \bigcap_{\alpha \in \emptyset} A_\alpha \ \Longleftrightarrow \exists\alpha: \ \alpha\in \emptyset\ \wedge\ x\notin A_\alpha $

Answer 4

Note that actually, $\cap_{i\in \emptyset} A_i = S$ is not a convention, it's actually just wrong.

The correct statement would be $\cap_{i\in \emptyset} A_i = \emptyset$, or $\bigwedge_{i\in \emptyset} A_i = S$, where $\bigwedge$ denotes the greatest lower bound in the complete lattice $(\mathcal{P}(S), \subset)$.

Why are these true ? Well first note that $\cap_{i\in \emptyset} A_i = S$ simply cannot be true. One the LHS we have an intersection of sets with no mention of $S$ (the $A_i$'s are subsets of $S$, but they're also subsets of $S\cup \{S\}$ or god knows what), and on the RHS we have a set $S$, that happens to contain the $A_i$'s. So thats just wrong because then, $\cap_{i\in \emptyset} A_i = S\cup\{S\}$ would also be true (the $A_i$s are subsets of $S\cup\{S\}$, after all), which is absurd.

So $\bigcap_{i\in I}A_i$ is actually an abbreviation for $\bigcap \{A_i \mid i\in I\}$ which is by definition $\{x\in \displaystyle\bigcup_{i\in I} A_i \mid \forall i\in I, x\in A_i\}$, where $\displaystyle\bigcup_{i\in I}A_i$ is an abbreviation for $\bigcup\{A_i\mid i\in I\}$ which is $\{x\mid \exists i\in I, x\in A_i\}$ (its existence is provided by the axiom of reunion). But then if $I=\emptyset$, this last thing is empty, and so are all its subsets, in particular $\cap_{i\in \emptyset} A_i $, which must then be empty.

However, in the aforementioned complette lattice, $\cap$ and $\wedge$ coincide on all sets but the emptyset. It is easy to see that in a complete lattice, the greatest lower bound of the emptyset is the lattice's maximum element. Therefore $\bigwedge_{i\in \emptyset}A_i = \bigwedge\emptyset = S$. Note that here, $\bigwedge$ does depend on $S$ so this makes complete sense.

However, in practice, since $\bigwedge$ and $\cap$ coincide so often, we ignore those subtleties and declare them to be equal, which leads to the abuse of notation $\bigcap_{i\in \emptyset}A_i = S$, which is what the other answers essentially tell you. But this is an abuse of notation (by the usual definitions etc.)

Answer 5

Those conventions comes from the fact that $(\mathcal P (S), \cup, \emptyset)$ and $(\mathcal P (S), \cap, S)$ are monoids.

Hence, those conventions are just the usual $\sum_{k \in \emptyset} k = 0$ or $\prod_{k \in \emptyset} k = 1$ that you certainly know for $(\mathbb N, +)$ or $(\mathbb Z, \times)$ for example.

Answer 6

But if "$\bigcap_{\alpha \in \emptyset} A_\alpha = S$" then "$x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha \in \emptyset$", meaning then $x \in \bigcup_{\alpha \in I}A_\alpha=S$.

Can someone help me and tell me what is wrong with this? Thank you.

This is a bit late, but recall that when dealing with an empty domain, universality does not imply existence.

$$\forall \alpha\in\emptyset ~(P(\alpha)) ~\nvdash~ \exists \alpha\in\emptyset~(P(\alpha))$$

And the definition of the empty intersection is: $~\bigcap_{\alpha\in\emptyset} A_\alpha~{=\{x\in S:\forall \alpha\in\emptyset~(x\in A_\alpha)\}\\=\{x\in S:\top\}\\=S}$

While the definition of the empty union is: $\qquad\bigcup_{\alpha\in\emptyset} A_\alpha~{=\{x\in S:\exists \alpha\in\emptyset~(x\in A_\alpha)\}\\=\{x\in S:\bot\}\\=\emptyset}$

Answer 7

$(x ∈ A_{\alpha} , \forall \alpha \in I )$ does not imply $( \exists \alpha \in I \text{ such that } x \in A_{\alpha} )$, since I may be $\emptyset$