Am I allowed to use distributive law for infinitely many sets?

Solution 1:

Yes, both distributive laws generalize. (You need only one of the two, but it’s useful to know both.) The first step in verifying the generalization that you want is to check that

$$\left(\bigcup_{i\in I}A_i\right)\cap D=\bigcup_{i\in I}(A_i\cap D)\;,\tag{1}$$

and to verify its mate you’ll want to check that

$$\left(\bigcap_{i\in I}A_i\right)\cup D=\bigcap_{i\in I}(A_i\cup D)\;.\tag{2}$$

Both are easily verified by element-chasing. For $(1)$, if $x\in\left(\bigcup_{i\in I}A_k\right)\cap D$, then $x\in\bigcup_{i\in I}A_i$ and $x\in D$. Since $x\in\bigcup_{i\in I}A_i$, there is an $i_0\in I$ such that $x\in A_{i_0}$, and therefore $x\in A_{i_0}\cap D\subseteq\bigcup_{i\in I}(A_i\cap D)$. Conversely, if $x\in\bigcup_{i\in I}(A_i\cap D)$, then there is an $i_0\in I$ such that $x\in A_{i_0}\cap D$. Then $x\in A_{i_0}\subseteq\bigcup_{i\in I}A_i$, and $x\in D$, so $x\in\left(\bigcup_{i\in I}A_i\right)\cap D$.

For $(2)$, if $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$, then $x\in\bigcap_{i\in I}A_i$ or $x\in D$. Let $i_0\in I$ be arbitrary. Then $A_{i_0}\supseteq\bigcap_{i\in I}A_i$, so $x\in A_{i_0}$ or $x\in D$, and therefoer $x\in A_{i_0}\cup D$. Since this holds for each $i_0\in I$, $x\in\bigcap_{i\in I}(A_i\cup D)$. Conversely, if $x\in\bigcap_{i\in I}(A_i\cup D)$, then $x\in A_i\cup D$ for each $i\in I$. If $x\in D$, then certainly $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$. If $x\notin D$, then we must have $x\in A_i$ for each $i\in I$, in which case $x\in\bigcap_{i\in I}A_i\subseteq\left(\bigcap_{i\in I}A_i\right)\cup D$.

Two applications of $(1)$ will give you the distributive law that you want. Suppose that $C=\bigcup_{i\in I}A_i$ and $D=\bigcup_{j\in J}B_j$. Then

$$\begin{align*} C\cap D&=\left(\bigcup_{i\in I}A_i\right)\cap D\\ &=\bigcup_{i\in I}(A_i\cap D)\\ &=\bigcup_{i\in I}\left(A_i\cap\bigcup_{j\in J}B_j\right)\\ &=\bigcup_{i\in I}\left(\bigcup_{j\in J}(A_i\cap B_j)\right)\\ &=\bigcup_{\langle i,j\rangle\in I\times J}(A_i\cap B_j)\;. \end{align*}$$

In other words, $C\cap D$ is the union of all possible intersections of the form $A_i\cap B_j$. In particular, if $I$ and $J$ are countable index sets, $I\times J$ is also countable.

Similarly, two applications of $(2)$ will give you the other general distributive law of this kind. This time suppose that $C=\bigcap_{i\in I}A_i$ and $D=\bigcap_{j\in J}B_j$. Then

$$\begin{align*} C\cup D&=\left(\bigcap_{i\in I}A_i\right)\cup D\\ &=\bigcap_{i\in I}(A_i\cup D)\\ &=\bigcap_{i\in I}\left(A_i\cup\bigcap_{j\in J}B_j\right)\\ &=\bigcap_{i\in I}\left(\bigcap_{j\in J}(A_i\cup B_j)\right)\\ &=\bigcap_{\langle i,j\rangle\in I\times J}(A_i\cup B_j)\; \end{align*}$$

the intersection of all possible unions of the form $A_i\cup B_j$ as $i$ and $j$ run over their respective index sets.

Solution 2:

Two sets $X$ and $Y$ are equal exactly when, for all $x$, $x \in X \iff x \in Y$.

Now, take $x \in (\cup_i C_i) \cap (\cup_j C_j)$.

This means precisely that $x$ is in $\cup_i C_i$ and also is in $\cup_j D_j$.

A statement of the form $x \in \cup_k S_k$ means precisely "there exists some $k$ such that $x \in S_k$". So we have from $x \in \cup_i C_i$ and $x \in \cup_j D_j$ that there exist $i$ and $j$ such that $x \in C_i$ and $x \in D_j$.

Or in other words, there exist $i$ and $j$ such that $x \in C_i \cap D_j$.

Or in other words, $x \in \cup_i \cup_j (C_i \cap D_j)$.

So, yes, the distributive law works for infinite, even uncountable families of sets.