Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like some help understanding this definition:

Definition. A directed family of subsets of a set $S$ is a family $(D_i)_{i\in I}$ of subsets of $S$ such that, for every $i$,$j$ $\in I$ there is some $k \in I$ such that $D_i \subseteq D_k$ and $D_j\subseteq D_k$.

Taken from from Grillet's Abstract Algebra

First of all, what is $I$ to which $i$ and $j$ belong? I get that if we have the subsets $ A_1, A_2, A_3 $ then $I = \{ 1,2,3 \}$ but I can't find any formal definition in the text. I don't get it in the more general form above.

Secondly, why do we use both $i$ and $j$ from $I$? Isn't it enough to say that for every $i$, we have a $k$ such that $D_i \subseteq D_k$? I'm trying to understand this as some kind of chain. What is the meaning of using both $i$ and $j$? If we have $D_i \subseteq D_k$ and $D_j\subseteq D_k$ does it say anything about how $D_i$ relate to $ D_j$?

(and exactly what does $(D_i)_{i\in I}$ mean in terms of notation?)

Could someone perhaps clarify this a bit for me. I'm rather confused as you might notice. Any help would be highly appreciated.

share|improve this question

2 Answers 2

  1. Usually, $I$ denotes a general index set. It may be finite (like in your example), countably infinite (then often $\Bbb{N}$ is used) or even uncountably infinite.
  2. Two indices are used, because you don't necessarily have one chain, like you mentioned. Take for example the sets $A = \{1,2\}$ and $B=\{3,4\}$. You can't compare them directly with $\subseteq$, but you have a bigger set, let's say $C = \{1,2,3,4,5\}$, that contains both of them.
  3. $(D_i)_{i\in I}$ denotes a family of sets, these are indexed by the set $I$.
share|improve this answer

You can think of $I$ as an index set. As you have mentioned if $I = \{ 1,2,3\}$ then $I \subset \mathbb{N}$. Now it is not necessary that $I$ is finite. It may be $I$ is infinite, possibly uncountable. This brings us to the second part of your question.

Secondly, why do we use both i and j from I? Isn't it enough to say that for every i, we have a k such that $D_i⊆D_k$? I'm trying to understand this as some kind of chain. What is the meaning of using both i and j? If we have $D_i \subseteq D_k$ and $D_j \subseteq D_k$ does it say anything about how Di relate to Dj?

You can think of $I$ in a more general setting where $I$ is partially ordered. If a set is partially ordered then it is possible the two elements are not comparable. For example, Let $I = \{Y,U,V\}$ such that $ U \subset Y$ and $ V \subset Y\setminus U$. If we define the ordering of the set by reverse inclusion, ie. $Y \leq U$ if $U \subset Y$. Then $ Y \leq U$ and $ Y \leq V$ but we cannot say $ U \leq V$ or $ V \leq U$. So, they are "incomparable" according to the "rules" we have used to define the index set.

(and exactly what does $(D_i)_{i \in I}$ mean in terms of notation?)

Now imagine that you have set $D$ and you would like to index elements of $D$. When the number of elements in your sequence becomes very large it is quiet cumbersome to write them out. Assuming the you are indexing over natural numbers, $(D_i)_{i \in \mathbb{N}}$ is a short hand notation of saying all elements of $\{D_1,\ldots D_i,D_{i+i},\ldots\}$ From the example above, if we set the index set of $D$ as $I$, a possible sequence could be $D_y,D_u$ as $ y \leq u$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.