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Theorem: Given $\{E_{\alpha}\}_{\alpha \in \mathcal{A}}$, where each $E_\alpha$ is a $\sigma$-algebra on $X$. Then $E:=\bigcap_{\alpha \in \mathcal{A}}E_\alpha$ is a $\sigma$-algebra.

Proof: Take $\{S_i\}_{i=1}^\infty$ with each $S_i \in E$. Then for each $\alpha \in \mathcal{A}$ we have $S_i \in E_\alpha$ and therefore $\bigcup_{i=1}^\infty S_i \in E_\alpha$. So $\bigcup_{i=1}^\infty S_i \in E$. And complement is same reasoning.

Now suppose we are given an arbitrary set $F$ and we say that $\{E_{\alpha}\}_{\alpha \in \mathcal{A}}$ is the set of all $\sigma$-algebras containing $F$. Then the above theorem would imply that there exists a unique smallest $\sigma$-algebra that contains $F$.

I don't know any set theory but I know that not all sets you can write down are legit (Russell's paradox), so I am a little uncomfortable with this. So I'd like to ask,

  1. Suppose we have $\{E_{\alpha}\}_{\alpha \in \mathcal{A}}$. That is, we have a function $f:\mathcal{A} \rightarrow \mathcal{P}\left(\mathcal{P}\left(X\right)\right)$, where $E_\alpha := f(\alpha)$. Can we always say that $\bigcap_{\alpha \in \mathcal{A}}E_\alpha$ is a legit set?

  2. Is it easy to show that you can construct $\{E_{\alpha}\}_{\alpha \in \mathcal{A}}$? (I.e. the existence of an $f$ such that $f(\alpha)=E_\alpha$) Do we need to worry about this constructability in practice? Why?

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up vote 4 down vote accepted

One of the standard axioms of set theory says that if $X$ is a set, then any definable collection of elements of $X$ is also a set. This is the axiom of separation or comprehension. "Definable" means here "expressible by a first order formula, with parameters".

In particular, $\bigcap_{\alpha\in\mathcal A}E_\alpha$ is a set as long as $\mathcal A\ne\emptyset$: Fix $\alpha_0\in\mathcal A$, and note that $$\bigcap_{\alpha\in\mathcal A}E_\alpha=\{x\in E_{\alpha_0}\mid\forall \alpha\in\mathcal A\,(x\in E_\alpha)\}.$$

The second issue is also fine: Each $E_\alpha$ is a $\sigma$-algebra on $X$, so it is a collection of subsets of $X$, so it is a subset of $\mathcal P(X)$, so it belongs to $\mathcal P(\mathcal P(X))$. The function $f$ has domain a set, in this case $D=\mathcal A$, and range $R$ contained in a set, in this case the double power set of $X$. In set theory we code functions by their graphs, so $f$ is just a subset of $D\times R$ with some properties, which means (again) that it is a set by separation. That $D\times R$ itself is a set also follows from standard axioms as long as $D$ and $R$ are sets; the key is that we can code ordered pairs as sets.

In general, we do not encounter any foundational issues at this level: In order to obtain a collection that is not a set, for example by taking a union $\bigcup_{i\in I}A_i$ or a product $\prod_{i\in I}A_i$, we would need that either the index set $I$ or one of the $A_i$ is not a set. We simply do not encounter these issues in practice. The axioms of set theory are powerful enough that imply that we obtain sets if we begin with sets, and apply standard mathematical operations such as those obtained by iterating applications of pairing $x,y\mapsto\{x,y\}$, taking unions, power-sets, forming subcollections of given sets, or taking pointwise images of sets.

(On the other hand, it is easy to describe collections or objects that are not sets: The collection of all groups, or the collection of all $\sigma$-algebras.)

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Note that Russell's paradox comes from trying to collect all objects with a given property (in this case, all objects $x$ such that $x\notin x$). Separation, on the other hand, is formulated precisely to avoid this issue, since it can only collect together all such objects that already belong to a set. This is the so-called Cantorian limitation of size, and is central to standard set theory, $\mathsf{ZF}$, see – Andrés E. Caicedo Apr 28 '13 at 6:18
Other set theories follow other conventions; if you are interested in this, you may want to look at this question: – Andrés E. Caicedo Apr 28 '13 at 6:19
How do we know that $D=\mathcal{A}$ is a set? Do we just take $\mathcal{A} = \{ \text{the set of all } \sigma \text{-algebras containing } F \}$, and then define $f=\{ (x,x) : x\in D \}$? So why do we even use indexed sets, it seems we could just write $\bigcap_{C\in \mathcal{A}} C$? – StuartHa Apr 28 '13 at 16:26
@Stuart Sure, we could even write $\bigcap \mathcal A$. But we use indexed sets because sometimes it is more natural to use a different index set than the set itself we are considering. For example, we could have several families, all indexed by the same set, so even though it is not mathematically necessary, it tends to be notationally advantageous. Anyway, if $\mathcal A$ is not a set, the intersection is still a set, as it is still given by the separation as before. – Andrés E. Caicedo Apr 28 '13 at 17:53
(Of course, if $\mathcal A$ is not a set, then mentions of $\mathcal A$ are understood as abbreviations where $\mathcal A$ is replaced by a description of what it is. We call such collections $\mathcal A$ proper classes. That said, in your setting, since the collection of $\sigma$-algebras over a given set is a set anyway, then, even if $\mathcal A$ is a proper class, there is a set $\mathcal A'$ that we can use instead giving rise to precisely the same intersection.) – Andrés E. Caicedo Apr 28 '13 at 17:54

A $\sigma$-algebra on a set $X$ is a subset of $\mathcal{P} ( X )$, meaning that they are elements of $\mathcal{P} ( \mathcal{P} ( X ) )$. The usual axioms of set-theory imply that the collection $$\mathfrak{S} = \{ \mathcal{S} \in \mathcal{P} ( \mathcal{P} ( X ) ) : \mathcal{S}\text{ is a }\sigma\text{-algebra on }X \}$$ is a set, and so are any subcollections of this set.

The paradoxes of naive set theory came about mainly because the "size" of the collections considered were unbounded; e.g., the family of all sets, the family of all ordinal numbers, etc. As long as you restrict yourself to considering collections that are built from some specific set $X$ in a definite way (using, for example, power set and subset definition) no paradoxes have ever come around.

Note that if $\{ \mathcal{S}_i : i \in I \} \subseteq \mathfrak{S}$ is nonempty, then picking any $i_0 \in I$ we have that $$\bigcap_{i \in I} \mathcal{S}_i = \{ A \in \mathcal{S}_{i_0} : ( \forall i \in I ) ( A \in \mathcal{S}_i ) \}$$ and so this causes no problems since we are using a subset definition of a "good" set.

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