Does Indexed Union distribute over Indexed Intersection? (same index)

Question

I am wondering whether this is true:

$$ \big(\bigcup_i A_i\big)\ \cap\ \big(\bigcup_i B_i\big) = \bigcup_i \big( A_i \cap B_i \big) $$

I have found proof that union distributes over intersection of not indexed sets and some proof where only one of the two sets is indexed, but I cannot prove myself whether this holds or not.

Any ideas?

Answer 1

It doesn't hold, not even for finite index sets.

For example

$$(\{1\} \cup \{2\}) \cap (\{2\} \cup \{1\}) = \{1,2\}$$ but $$(\{1\} \cap \{2\}) \cup (\{2\} \cap \{1\}) = \emptyset$$

What is true is this:

$$\left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) = \bigcup_{i\in I, j \in J} A_i \cap B_j$$

Answer 2

A proof of the statement that mechanodroid has indicated to be true:

Let $X$ be a nonempty set and let $\{A_i : i \in I \}, \{ B_j : j \in J\} \subset P(X)$ (to avoid trivialities, we also assume that there is a nonempty set in each of these indexed families of sets). We will show that \begin{equation} \left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) = \bigcup_{i\in I,\, j \in J} A_i \cap B_j . \end{equation}

Suppose $x \in \left(\bigcup_{i\in I} A_i \right) \cap \left(\bigcup_{j\in J} B_j\right)$. So $x \in \bigcup_{i\in I} A_i$ and $x \in \bigcup_{j\in J} B_j$ (definition of set intersection). By the definition of set union for an indexed family of sets, there exists $k \in I$ and there exists $l \in J$ such that $x \in A_k$ and $x \in B_l$. Hence $x \in A_k \cap B_l$. Thus, $\left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) \subseteq \bigcup_{i\in I,\, j \in J} A_i \cap B_j $, since clearly $x \in \bigcup_{i\in I,\, j \in J} A_i \cap B_j \left( \supseteq A_k \cap B_l \right)$.

Suppose $x \in \bigcup_{i\in I,\, j \in J} A_i \cap B_j $. Then there exists $(k,l) \in I \times J$ such that $x \in A_k \cap B_l$. So there exists $k \in I$ and $l \in J$ (definition of the Cartesian product) such that $x \in A_k$ and $x \in B_l$. Thus we have that $x \in \bigcup_{i\in I} A_i$ and $x \in \bigcup_{j\in J} B_j$, and so $x \in \left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right)$. Therefore, $\left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) = \bigcup_{i\in I,\, j \in J} A_i \cap B_j $.

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As a counterexample (to the claim in aky's question) with a countably infinite indexing set ($\mathbb{N}$) and sets that contain more than one point (for fun, since mechanodroid's answer is already sufficient, as well as a stronger counterexample):

For $n=1,2, \ldots$ we define the following subsets of $\mathbb{R}$

\begin{align} & A_n:=\Big[-1, \frac{1}{n}-\frac{1}{2} \Big] \text{ , and} \\& B_n:=\Big[\frac{1}{n} , \frac{3}{2} \Big] . \end{align} So \begin{align}\left(\bigcup_{n\in \mathbb{N}} A_n\right) \bigcap \left(\bigcup_{n\in \mathbb{N}} B_n\right) & = [-1,1/2 ] \cap (0, 3/2] \\& = (0, 1/2], \end{align}

BUT

\begin{align}\bigcup_{n\in \mathbb{N}} \left( A_n \cap B_n \right) & = \bigcup_{n\in \mathbb{N}} \emptyset \\& = \emptyset. \end{align}