Understanding Set Notation

Question

I'm having some trouble understanding a definition and explanation in my textbook Introduction to Analysis by Edward Gaughan 5th edition. The book begins with some preliminary information about sets and some of the information is sort of confusing and I have a feeling that in this sort of class if you don't understand the basics you are doomed. So hopefully somebody out there can help me understand the confusing bits a little better.

So the definition in question is as follows:

Let $\Lambda$ be a set, and suppose for each $\lambda \in \Lambda$, a subset $A_{\lambda}$ of a given set S is specified. The collection of sets $A_{\lambda}$ is called an indexed family of subsets of S with $\Lambda$ as the index set. We denote this by ${\{ A \}}_{\lambda \in \Lambda}$

If ${\{ A \}}_{\lambda \in \Lambda}$ is an indexed family of sets, define

$\cap_{\lambda \in \Lambda}$ $A_{\lambda}$ = {x : x $\in A_{\lambda}$ for all $\lambda \in \Lambda$}

$\cup_{\lambda \in \Lambda}$ $A_{\lambda}$ = {x : x $\in A_{\lambda}$ for some $\lambda \in \Lambda$}

Alright so the definition above I thought was straightforward, but then the text goes on to say this:

There is one logical difficulty here that needs to be pointed out. If $\Lambda$ is the empty set, then it is easy to see that $\cup_{\lambda \in \Lambda} A_{\lambda}$ is empty; however, it is not clear what to expect of $\cap_{\lambda \in \Lambda} A_{\lambda}$.

The bolded explanation is what I do not understand. If $\Lambda$ is the empty set there will be no subsets to union because there are no elements $\lambda \in \Lambda$ but why is it not clear that $\cap_{\lambda \in \Lambda} A_{\lambda}$ is also empty, won't there also be no subsets $A_{\lambda}$ to seek the intersection of??

The text ends with the following conclusion:

In a context where all sets considered are understood to be subsets of a given set S, the common usage is to let:

S = $\cap_{\lambda \in \oslash} A_{\lambda}$

Any information would be appreciated.

Answer 1

By definition $x\in\bigcap_{\lambda\in\Lambda}A_\lambda$ if and only if $x\in A_\lambda$ for all $\lambda\in\Lambda$. Turn that around: what does it mean to say that $x\notin\bigcap_{\lambda\in\Lambda}A_\lambda$? It must mean that there is some $\lambda\in\Lambda$ such that $x\notin A_\lambda$. To put it a little differently, suppose that someone gave you an $x$ and asked you to prove that $x\notin\bigcap_{\lambda\in\Lambda}A_\lambda$; how could you do it? The only sure way seems to be to find some $\lambda\in\Lambda$ such that $x\notin A_\lambda$. But if $\Lambda=\varnothing$, you can’t even find a $\lambda\in\Lambda$, let alone one with the further property that $x\notin A_\lambda$. Thus, when $\Lambda=\varnothing$ there is no way to show that some $x\notin\bigcap_{\lambda\in\Lambda}A_\lambda$!

On the face of it this would seem to show that every $x$ belongs to $\bigcap_{\lambda\in\Lambda}A_\lambda$. Unfortunately, that ultimately leads to Russell’s paradox, so we can’t allow it. The best we can do is take $\bigcap_{\lambda\in\Lambda}A_\lambda$ to be the set of all objects in the universe of discourse, Gaughan’s set $S$.

Answer 2

If we have the index set being empty, then $$ \bigcap_{\lambda \in \Lambda} A_\lambda = \{x : x \in A_\lambda\text{ for all }\lambda \in \emptyset \}.$$

The problem here is that the statement $x \in A_\lambda$ for all $\lambda \in \emptyset$ is vacuously true, and so all x satisfy it. The question then becomes, what do we mean by all $x$? This is why it's helpful to establish some superset that we assume we're working in. This super set is the S referred to in the book, and this takes the place of all $x$.

Answer 3

We have $x\in \bigcup A_\lambda$ iff there exists a $\lambda\in \Lambda$ with $x\in A_\lambda$. If there exists no $\lambda$ at all, there certainly exists no lambda with $x\in A_\lambda$. Hence $x\in\bigcup A_\lambda$ is false for whatever $x$ we check. Hence the set contains nothing, it is the empty set.

We should have $x\in \bigcap A_\lambda$ iff for all $\lambda\in \Lambda$ we have $x\in A_\lambda$. But since $\Lambda$ is empty, anything is true about the elements of $\Lambda$, e.g. each element of $\Lambda$ is a green-eyed fyling pig. Especially $x\in A_\lambda$ is true for all $\lambda$, hence $x\in\bigcap A_\lambda$ is true for whatever $x$ we check. Hence the set contains everyhing. However there is no such thing as a "set of everything". Well, there is a set of everything in limited constexts, as the conclusion of your text writes: If your objects of discourse are only the elements of $S$ then it is valid to say that "everything" refers to all elements of $S$, i.e. $S$ itself. As this notion of $\bigcap A_\lambda$ varies with $S$, there is no general notion of it for empty $\Lambda$.

Answer 4

Suppose that $\Lambda = \varnothing$. Let $x$ be a set. Then, for some $\lambda \in \Lambda$, we have $x \notin A_{\lambda}$. Since $\Lambda = \varnothing$ there is no such $\lambda$. Thus $x \in \cap \{ A_{\lambda} \colon \lambda \in \varnothing \} $. Thus $\cap \{ A_{\lambda} \colon \lambda \in \varnothing \} $ has every set as an element. A more common approach to the empty union is to consider $A_{\lambda} \subseteq B$ for all $\lambda$ and let $\cap$ be an operator on the complete lattice of subsets of $B$. In this case the $x$ we used above is required to be an element of $B$. The empty intersection will be $B$. In practice the $A_{\lambda}$ are coming from somewhere and $B$ is a set in question that contains all of the $A_{\lambda}$.