This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Family of sets
Let an index set \(I\) and a set of sets \(\mathcal{S}\) be given. Then a function \(f=(F,I,\mathcal{S})\) from \(I\) to \(\mathcal{S}\) assigns to each \(i\in I\) an element of \(\mathcal{S}\), i.e., a set. Previously we agreed to write this as \((f_i)_{i\in I}\), but to maintain the convention of writing sets with capital letters, we shall write it as \((F_i)_{i\in I}\).
Suppose every set in the family \((A_i)_{i\in I}\) is a subset of some set \(A\). Then we may take the target \(\mathcal{S}\) of this function to be \(\mathcal{P}(A)\). When we wish to regard the sets \(A_i\) as subsets of \(A\) in this way, we say that \((A_i)_{i\in I}\) is a family of subsets of the set \(A\).
Union and intersection
In §ZFC Axioms we accepted as an axiom that unions exist. The following notation is used somewhat more frequently than the notation we employed when introducing that axiom.
Definition 1 Let \((A_i)_{i\in I}\) be a family of sets. Then the set of all \(x\) that
Thus the union is the set of all \(x\) satisfying the logical formula
\[\exists i(i\in I\wedge x\in A_i)\]Hence if \(I=\emptyset\), then \(\bigcup_{i\in I} A_i=\emptyset\). It is not difficult to verify that the union does not depend on the target \(\mathcal{S}\) of \((A_i)_{i\in I}\).
Definition 2 Let \((A_i)_{i\in I}\) be a family of sets, and suppose \(I\) is not empty. Then the set of all \(x\) that
The intersection is the collection of all \(x\) satisfying the logical formula
\[\forall i(i\in I\implies x\in A_i)\]If \(I=\emptyset\), then \(i\in I\) is false, so the entire statement is true regardless of \(x\), and \(\bigcap_{i\in\emptyset} A_i\) would have to be the universal set, which is a contradiction. (§ZFC Axioms, ⁋Example 4) If we specify the target \(\mathcal{S}\) of \((A_i)_{i\in I}\) appropriately, we can define the intersection while avoiding this contradiction.
Definition 3 Let \((A_i)_{i\in I}\) be a family of subsets of a set \(A\). Then the set of all \(x\) that
This time, even if \(I=\emptyset\), the condition becomes
\[(x\in A)\wedge (\forall i(i\in I\implies x\in A_i))\]so that \(\bigcap_{i\in I} A_i=A\). Henceforth, whenever we take the intersection of a family of sets in any proposition, we assume either that \(I\) is nonempty or that the given family is a family of subsets of some set.
Proposition 4 Consider a family of sets \((A_i)_{i\in I}\) and a surjective function \(f:K\rightarrow I\). Then the two equalities
\[\bigcup_{k\in K}A_{f(k)}=\bigcup_{i\in I}A_i,\qquad \bigcap_{k\in K}A_{f(k)}=\bigcap_{i\in I}A_i\]hold.
Proof
First suppose \(x\in\bigcup_{i\in I} A_i\). That is, for some \(i_0\in I\) we have \(x\in A_{i_0}\). Since \(f\) is surjective, there exists some \(k_0\in K\) such that \(i_0=f(k_0)\), and therefore \(x\in A_{f(k_0)}\), so \(x\in\bigcup_{k\in K}A_{f(k)}\).
Conversely, if \(x\in\bigcup_{k\in K}A_{f(k)}\) holds, then for some \(k_0\in K\) we have \(x\in A_{f(k_0)}\). Since \(f(k_0)\in I\), the set \(A_{f(k_0)}\) is one of the sets constituting \((A_i)_{i\in I}\), and therefore \(x\in \bigcup_{i\in I} A_{i}\).
We must now show the second equality. First suppose \(x\in\bigcap_{i\in I}A_i\). Then \(x\in A_i\) for every \(i\in I\). For arbitrary \(k_0\in K\) we have \(f(k_0)\in I\), so \(x\in A_{f(k)}\) for every \(k\in K\), and therefore \(x\in \bigcap_{k\in K}A_{f(k)}\). Conversely, if \(x\in A_{f(k)}\) for every \(k\in K\), then since \(f\) is surjective, \(x\in A_i\) also holds for every \(i\in I\).
In particular, suppose \((A_k)_{k\in K}\) satisfies \(A_k=A_{k'}\) for arbitrary \(k,k'\in K\). Choose any \(k_0\in K\), let \(I=\{k_0\}\), and apply the above proposition to the surjective function \(f:K\rightarrow I\); then since \(A_k=A_{k_0}=A_{f(k)}\) for every \(k\),
\[\bigcup_{k\in K} A_k=\bigcup_{k\in K} A_{f(k)}=\bigcup_{i\in I}A_i=A_{k_0},\qquad \bigcap_{k\in K}A_k=\bigcap_{k\in K}A_{f(k)}=\bigcap_{i\in I}A_i=A_{k_0}\]hold.
Let us now examine a few more properties of unions and intersections. If two families \((A_i)_{i\in I}\) and \((B_i)_{i\in I}\) with the same index set are given, and \(B_i\subseteq A_i\) holds for every \(i\in I\), then
\[\bigcup_{i\in I} B_i\subset\bigcup_{i\in I} A_i,\qquad \bigcap_{i\in I} B_i\subset\bigcap_{i\in I} A_i\]is obvious. Also, for a given family \((A_i)_{i\in I}\) and a subset \(J\) of \(I\),
\[\bigcup_{j\in J}A_j\subset\bigcup_{i\in I} A_i,\qquad\bigcap_{j\in J}A_j\supset\bigcap_{i\in I} A_i\]is almost obvious.
Associativity
The operations on sets satisfy associativity.
Proposition 5 Let \((A_i)_{i\in I}\) be a family of sets, and suppose the index set \(I\) is the union of a family \((J_k)_{k\in K}\). Then
\[\bigcup_{i\in I} A_i=\bigcup_{k\in K}\left(\bigcup_{j\in J_k} A_j\right),\quad \bigcap_{i\in I}A_i=\bigcap_{k\in K}\left(\bigcap_{j\in J_k} A_j\right)\]hold.
Proof
Let us first prove the equality for unions. If \(x\in \bigcup_{i\in I}A_i\), then for some \(i_0\in I\) we have \(x\in A_{i_0}\). Since \(I=\bigcup_{k\in K} J_k\), there exists some \(k_0\) such that \(i_0\in J_{k_0}\). Then
\[A_{i_0}=\bigcup_{i\in \{i_0\}}A_i\subset\bigcup_{j\in J_{k_0}} A_j=\bigcup_{k\in\left\{k_0\right\}}\left(\bigcup_{i\in J_k} A_i\right)\subseteq \bigcup_{k\in K}\left(\bigcup_{j\in J_k} A_j\right)\]so \(x\in A_{i_0}\subseteq \bigcup_{k\in K}\left(\bigcup_{j\in J_k} A_j\right)\).
Conversely, if \(x\in \bigcup_{k\in K}\left(\bigcup_{j\in J_k} A_j\right)\), then for some \(k_0\in K\) we have \(x\in \bigcup_{j\in J_{k_0}}A_j\), and thus for some \(i_0\in J_{k_0}\) we have \(x\in A_{i_0}\). Since \(i_0\in I\), we obtain \(x\in\bigcup_{i\in I} A_i\).
The second equality can be proved similarly. If \(x\in\bigcap_{i\in I} A_i\), then \(x\in A_i\) for every \(i\in I\). For arbitrary \(k\in K\) we have \(J_{k}\subseteq I\), so the statement that the above holds for every \(i\in I\) is equivalent to the statement that \(x\in A_j\) holds for every \(j\in J_{k}\). Since this holds for arbitrarily chosen \(k\), this means exactly that \(x\in\bigcap_{k\in K}\left(\bigcap_{j\in J_{k}}A_j\right)\).
Image of a union or intersection
We may also consider the image of a union or intersection as follows.
Proposition 6 Let \((A_i)_{i\in I}\) be a family of subsets of a set \(A\), and let \((R,A,B)\) be a binary relation. Then
\[R\left(\bigcup_{i\in I} A_i\right)=\bigcup_{i\in I}R(A_i),\quad R\left(\bigcap_{i\in I} A_i\right)\subset\bigcap_{i\in I}R(A_i)\]Proof
Let us first prove the first equality. If \(y\in R\left(\bigcup_{i\in I}A_i\right)\), then there exists suitable \(x\in \bigcup_{i\in I}A_i\) such that \((x,y)\in R\). Now if \(x\in A_j\), then \(y\in R(A_j)\), so \(y\in\bigcup_{i\in I}R\left(A_i\right)\) holds. Conversely, if \(y\in \bigcup_{i\in I}R\left(A_i\right)\), then for some \(j\) we have \(y\in R\left(A_j\right)\), so there exists suitable \(x\in A_j\) such that \((x,y)\in R\). Therefore \(y\in R\left(\bigcup_{i\in I} A_i\right)\) holds.
For the second equality it suffices to show one inclusion. Suppose \(y\in R\left(\bigcap_{i\in I}A_i\right)\). Then there exists some \(x\in\bigcap_{i\in I}A_i\) such that \((x,y)\in R\). Since \(x\) belongs to every \(A_i\), we know that \((x,y)\in R(A_i)\) holds for every \(A_i\). That is, \(y\in \bigcap_{i\in I}R\left(A_i\right)\).
The reverse inclusion in the second equality of the above proposition does not hold in general, but if \(R\) is the inverse relation of a function, then equality holds.
Proposition 7 Let \(f:A\rightarrow B\) be a function, and let \((B_i)_{i\in I}\) be a family of subsets of \(B\). Then
\[f^{-1}\left(\bigcap_{i\in I} B_i\right)=\bigcap_{i\in I} f^{-1}(B_i)\]holds.
Proof
One inclusion was proved in the more general case, so it suffices to prove the reverse inclusion.
Suppose arbitrary \(x\in\bigcap_{i\in I} f^{-1}(B_i)\) is given. Then \(x\in f^{-1}(B_i)\) for every \(i\). That is, for every \(i\) there exists \(y_i\in B_i\) such that \((x,y_i)\in F\). Since \(f\) is a function, such \(y_i\) is unique. Let this common value be \(y\); then \(y\in B_i\) for every \(i\in I\), so \(y\in\bigcap_{i\in I} B_i\), and therefore from \(f(x)=y\) we obtain \(x\in f^{-1}\left(\bigcap_{i\in I} B_i\right)\).
De Morgan’s laws
Proposition 8 (De Morgan’s law) For a family \((A_i)_{i\in I}\) of subsets of a set \(A\),
\[A\setminus \left(\bigcup_{i\in I}A_i\right)=\bigcap_{i\in I}(A\setminus A_i),\quad A\setminus\left(\bigcap_{i\in I} A_i\right)=\bigcup_{i\in I} (A\setminus A_i)\]hold.
Proof
To prove the first equality, suppose first \(x\in A\setminus\left(\bigcup_{i\in I} A_i\right)\). Then \(x\in A\) and \(x\not\in\left(\bigcup_{i\in I} A_i\right)\). Hence \(x\not\in A_i\) for every \(i\), so \(x\in (A\setminus A_i)\) holds for every \(i\). That is, \(x\in\bigcap_{i\in I}(A\setminus A_i)\). Conversely, if \(x\in\bigcap_{i\in I} (A\setminus A_i)\), then for arbitrary \(i\in I\) we have \(x\in A\setminus A_i\), and therefore \(x\not\in A_i\) for every \(i\in I\). Now since \(x\not\in\bigcup_{i\in I} A_i\), we have \(x\in A\setminus\bigcup_{i\in I} A_i\).
The second equality is obvious from the first, since the equality \(A\setminus(A\setminus X)=X\) holds for every \(X\subseteq A\).
References
[Bou] N. Bourbaki, Theory of Sets. Elements of mathematics. Springer Berlin-Heidelberg, 2013.
댓글남기기