# When indexing set is empty, how come the union of an indexed family of subsets of S is an empty set?

And how is intersection of an indexed family of subsets if the set itself? I am reading the Bert Mendelson's Introduction to Topology, and I can't figure how this works. When the indexing set is an empty set, doesn't that mean we can't find any subset of the form $A_\alpha$? Therefore, there is no element in the set $S$ such that $x \in A_\alpha$? I could accept the idea that $\cup_{\alpha \in I}A_\alpha = \emptyset$, but can't understand why $\cap_{\alpha \in I}A_\alpha = S$. To my understanding, I think $\cap_{\alpha \in I}A_\alpha = \emptyset$ as well.

You have not specified what $S$ is. Anyway, if $S$ is assumed to be a set, then $\bigcap_\alpha A_\alpha\neq S$ since $\bigcap_\alpha A_\alpha$ is not a set, but it is a class which is merely a meta-language in ZFC. Let $x$ be a set. Suppose $x\notin \bigcap_\alpha A_\alpha$. Then, there exists $\alpha\in I$ such that $x\notin A_\alpha$, which is false since $I$ is nonempty. Hence $\bigcap_\alpha A_\alpha$ contains all sets, so it is not a set.

However, it is customary to write $\bigcap_\alpha A_\alpha$ be mean the whole space depending on context. For example, when $X$ is a topological space and $A_\alpha\subset X$, we define $\bigcap_\alpha A_\alpha$ to be $X$.

Here is a model you might find useful. It's not a formal argument, but an analogy important when thinking about writing computer programs that may help your intuition.

To find the sum S of a list L of numbers you write this code:

S = 0
while there are numbers left in L
let n be the next number in L
let S become S+n
remove n from L


It's easy to see that this will work in the expected case, when the list isn't empty. Good computer programs work even for boundary cases on their inputs. This one makes 0 the sum of an empty list, which makes intuitive sense.

If you wanted the product of the numbers in the list you would modify the code:

P = 1
while there are numbers left in L
let n be the next number in L
let P become P*n
remove n from L


For this to work you must set P to 1 before you start. Then it's clear but perhaps a little less intuitive that the product of an empty list should be 1.

When doing boolean algebra, unions are like sums and intersections are like products. To find the union of a list L of subsets of a given set X

U = empty
while there are subsets left in L
let A be the next subset in L
let U become U union A
remove A from L


For an empty list, the union is empty, as expected.

For the intersection of a list of subsets

I = X
while there are subsets left in L
let A be the next subset in L
let I become I intersect A
remove A from L


In the edge case when the list is empty the intersection is all of X. The whole set is the (multiplicative) identity for intersection just as the empty set is for union.

There are computer languages in which you can write all four of these algorithms as instances of the same pattern of computation.

I disagree with Hermes.

$x \in \bigcup_{i\in I} A_i$ iff $\exists i \in I$ s.t. $x\in A_i$. When $I = \emptyset$ there never exists such an $i$, then the RHS is false independent of what $x$ is, hence $x$ should not be in the empty union.

Similarly $x\in \bigcap_{i\in I} A_i$ iff $\forall i \in I[x\in A_i]$. When $I$ is empty, the RHS is vacuously true, and so everything is in the empty intersection.