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I'm puzzled by something that might be complete silly. Halmos writes in his "Naive Set Theory":

If $\mathcal C$ is a collection of subsets of a set $E$ (that is, $\mathcal C$ is a subcollection of $\mathcal P(E)$), then write

$$\mathfrak D = \{X \in \mathcal P(E):X^\prime \in \mathcal C\}$$

It is customary to denote the union and intersection of $\mathfrak D$ by the symbols:

$$\bigcup_{X \in \mathcal C}X' \;\;\text{ and } \;\;\bigcap_{X \in \mathcal C}X'$$

He is defining the union and intersection of the complements of the sets $X \in \mathcal C$, which are subsets of the set $E$. $E$ is the "everything" set, which contains all sets considered in the current section.

I'm a little confused by why he is considering the set $\mathfrak D$. Why not just consdier the set of complements? $\mathfrak D$ is apparently, as I understand, the set of all subsets of $E$ such that the complement of $X$, namely $X'$ is in the collection of subsets of $E$, $\mathcal C$. By definition of the powerset of $E$,

$$\mathcal C \subset \mathcal P(E)$$

so $\mathfrak D$ is the set of subsets of $E$, $X$ such that $X'$ is also a subset of $E$.

I still can't see how $\mathfrak D$ enters in the picture here.

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  • $\begingroup$ $\mathfrak D$ is the set of complements, but presumably in Halmos's formulation of set theory, you have you have only defined an intersection of a family of sets, and not the intersection of a family of things related to the sets (such as complement). $\endgroup$ – GEdgar Jul 14 '12 at 3:15
  • $\begingroup$ @GEdgar I understand what you mean. $\endgroup$ – Pedro Tamaroff Jul 14 '12 at 3:16
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$\mathfrak{D}=\{X\in\mathcal{P}(E):X'\in\mathcal{C}\}$ is the collection of subsets of $E$ whose complements are in $\mathcal{C}$. For instance, if $\mathcal{C}$ is the collection of all finite subsets of $E$, then $\mathfrak{D}$ is the collection of all cofinite subsets of $E$, i.e., those subsets of $E$ whose complements are finite.

Note that for subsets $X$ of $E$ we have $X\in\mathfrak{D}$ iff $X'\in\mathcal{C}$. Take complements: this is equivalent to saying that $X'\in\mathfrak{D}$ iff $X''\in\mathcal{C}$. But $X''=X$, so $X'\in\mathfrak{D}$ iff $X\in\mathcal{C}$, and $\mathfrak{D}=\{C':C\in\mathcal{C}\}$. Thus, we can just as well describe $\mathfrak{D}$ as the set of complements (relative to $E$) of members of $\mathcal{C}$. And this is what justifies writing

$$\bigcup\mathfrak{D}=\bigcup_{C\in\mathcal{C}}C'\quad\text{ and }\quad\bigcap\mathfrak{D}=\bigcap_{C\in\mathcal{C}}C'\;.\tag{1}$$

If it weren’t the case that $\mathfrak{D}=\{C':C\in\mathcal{C}\}$, $(1)$ would not be justifiable.

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  • $\begingroup$ Bah, what a farfetched way of putting it! Thanks! "Why not just consdier the set of complements?" $\endgroup$ – Pedro Tamaroff Jul 14 '12 at 3:14
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    $\begingroup$ @Peter: Yes, $\mathfrak{X}$ was a typo. I suspect that the form of the definition of $\mathfrak{D}$ was chosen because it’s the form required to justify forming the set using the axiom schema of comprehension. That is, you can’t form a set $\{x:\varphi(x)\}$, where $\varphi(x)$ is some statement about $x$; you can only form a set $\{x\in y:\varphi(x)\}$, where $y$ is already known to be a set. $\endgroup$ – Brian M. Scott Jul 14 '12 at 3:18
  • $\begingroup$ "Axiom schema of comprehension"$=$"Axiom of specification"? $\endgroup$ – Pedro Tamaroff Jul 14 '12 at 3:20
  • $\begingroup$ @Peter: Yes. I learnt it as comprehension, so that’s the term that I use naturally. $\endgroup$ – Brian M. Scott Jul 14 '12 at 3:23
  • $\begingroup$ Hehe, OK. Thank you Professor! $\endgroup$ – Pedro Tamaroff Jul 14 '12 at 3:24

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