I am currently trying to understand something that was stated in "Introduction to Analysis", it is the fifth edition. It is used for the Advance Calculus 1 course at ASU. I'm not taking the class right now but I want to get a jump on the material. I've taken an intro course on Logic and I've also taken a Discrete Mathematics course.

I think I understand these two definitions well. The first is "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 $\textit{indexed family}$ of subsets of S with $\Lambda$ as the index set. We denote this by $\{A_\lambda\}_{\lambda\in\Lambda}$".

The second is:
If $A_\lambda$ is an indexed family of sets, define
$$\bigcap_{\lambda\in\Lambda} A_\lambda=\{x:x \in A_\lambda, \text{for all }\lambda \in \Lambda\}$$
$$\bigcup_{\lambda\in\Lambda} A_\lambda=\{x:x \in A_\lambda, \text{for some } \lambda \in \Lambda\}$$
The book states that if $\Lambda$ is empty then the union will be the empty set but that it is unclear what to expect from the intersection. I don't understand why that is. If $\Lambda$ is empty then doesn't that mean that there is no index for $A_\lambda$ and that there is no way of creating the intersection and union of $A_\lambda$?


The union with empty index should be any set rather than empty. There are mainly 2 reasons for it. First, if all $A_{\lambda}$ are same, to be consistent with Idempotence of set the union should be $A_{\lambda}$ even if index is empty. More precisely, if for all $\lambda \in \Lambda, \: A_{\lambda}=A$, then $$ \bigcup_{\lambda\in\Lambda} A_\lambda=A $$ Thus union with empty index should be $$ \bigcup_{\lambda\in\Phi} A_\lambda=A $$ Second, if union with empty index is empty, then as you see, intersection would be universe by De Morgan's law, which makes no sense since intuitively union is always larger than intersection.

  • $\begingroup$ If I am understanding you correctly then the union should be any other set other than empty because if it is empty then it would lead to a contradiction because it would imply that the intersection is the universe? Also, would we not know what the union is with an empty $\Lambda$ where all $A_\lambda$ might not be the same? $\endgroup$ – gabrieldreal Mar 18 '16 at 18:02
  • $\begingroup$ Yes,your understanding is right. If they are not same, the union could be any set $\endgroup$ – Math Wizard Mar 18 '16 at 18:06

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