Union of a set of collections

Question

I'm given the following definition asked to prove the following theorem:

Definition: Let $X$ be a set and suppose $C$ is a collection of subsets of $X$. Then $\cup \mathbf{C}=\{x \in X : \exists C\in \mathbf{C}(x\in C)\}$

Theorem: Let $\mathbf{C,D}$ be collections of subsets of a set $X$. Prove that $\cup ( \mathbf{C} \cup \mathbf{D}) = (\cup \mathbf{C}) \cup (\cup\mathbf{D})$

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By my reading of the definition, I run into two problems:

Firstly I think there is a type error since $ ( \mathbf{C} \cup \mathbf{D}) = \cup\{\mathbf{C}, \mathbf{D}\}$ is not a collection of subsets of $X$ (i.e. it is not a set whose elements are subsets of $X$); instead it is a set whose elements are collections of subsets of $X$.

Second, if we ignore the type error and plug into the definition, we get: $\cup \{\mathbf{C}, \mathbf{D}\} =\{x\in X:\exists C\in \{\mathbf{C}, \mathbf{D}\}(x\in C)\}=\{x\in X:x\in \mathbf{C} \lor x \in \mathbf{D}\}$, however, since $\boldsymbol{C},\boldsymbol{D}$ are collections, all their elements are sets. Since $x$ is not a set, $x\notin\boldsymbol{C}\land x\notin\boldsymbol{D}$. Thus $\cup \{\mathbf{C}, \mathbf{D}\}=\emptyset$

However this can't be right since the other side of the equality; $(\cup \mathbf{C}) \cup (\cup\mathbf{D})\neq \emptyset$ in general. What am I missing?

Answer 1

Essential are the points:

• $a\in\cup b\iff \exists x[a\in x\in b]$
• $a\cup b$ is an abbreviation of $\cup\{a,b\}$ (so that $x\in\cup\{a,b\}\iff x\in a\vee x\in b$)

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On your first point:

$\mathbf C$ and $\mathbf D$ are collections of subsets of $X$.

Then $\{\mathbf C,\mathbf D\}$ is a set whose elements are collections of subsets of $X$.

Consequently $\cup\{\mathbf C,\mathbf D\}$ is a set whose elements are subsets of $X$, because: $$x\in\cup\{\mathbf C,\mathbf D\}\text{ iff }x\in\mathbf C\text{ or }x\in\mathbf D$$

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It must be proved that: $$\cup(\mathbf C\cup\mathbf D=(\cup\mathbf C)\cup(\cup\mathbf D)$$ or equivalently that: $$\cup\cup\{\mathbf C,\mathbf D\}=\cup\{\cup\mathbf C,\cup\mathbf D\}\tag1$$

Equivalent are the following statements:

• $x\in\cup\cup\{\mathbf C,\mathbf D\}$
• $\exists A\in\cup\{\mathbf C,\mathbf D\}[x\in A]$
• $\exists A[[A\in\mathbf C\vee A\in\mathbf D]\wedge x\in A]$
• $\exists A[x\in A\in\mathbf D]\vee\exists A[x\in A\in\mathbf D]$
• $x\in\cup\mathbf C\vee x\in\cup\mathbf D$
• $x\in\cup\{\cup\mathbf C,\cup\mathbf D\}$

This proves $(1)$

Answer 2

I think you're just confused by the definition. If $\mathbf{C}\subseteq 2^X$, then we write $\cup\mathbf{C}:=\bigcup_{C\in\mathbf{C}}C$. For the sake of clarity, let's use a different symbol. Let's instad write $\vee\mathbf{C}:=\bigcup_{C\in\mathbf{C}}C$. Now, when they ask you to prove $\cup(\mathbf{C}\cup\mathbf{D})=(\cup\mathbf{C})\cup(\cup\mathbf{D})$, in our new notation, they're really asking you to prove that $\vee(\mathbf{C}\cup\mathbf{D})=(\vee\mathbf{C})\cup(\vee\mathbf{D})$. What makes their notation confusing is that they're using the same symbol to mean two different things (however, this is standard notation).

Answer 3

The "binary" $\mathbf{C} \cup \mathbf{D}$, where $\mathbf{C,D} \subseteq \mathscr{P}(X)$ is just $\{A \subseteq X: A \in \mathbf{C} \text{ or } A \in \mathbf{D}\}$, and could indeed also have been written as $\bigcup \{\mathbf{C},\mathbf{D}\}$, and this would have made the left hand side of the equation you wanted to show equal to $\bigcup \bigcup \{ \mathbf{C},\mathbf{D}\}$ which looks a bit confusing.

To see the desired equality, two inclusions need to be shown, e.g.:

Let $x \in \bigcup \mathbf{C}\cup\mathbf{D}$ then there exists some $A \in \mathbf{C}\cup\mathbf{D}$ with $x \in A$. This $A$ is either in $\mathbf{C}$ or in $\mathbf{D}$. In the former case $x \in \bigcup \mathbf{C}$, in the latter $x \in \bigcup \mathbf{D}$, so $x \in (\bigcup \mathbf{C}) \cup (\bigcup \mathbf{D})$. The reverse inclusion is similar.

If we only use "non-binary" $\cup$, the asked for statement becomes

$$\bigcup \bigcup \{\mathbf{C}, \mathbf{D}\} = \bigcup\{\bigcup \mathbf{C}, \bigcup \mathbf{D} \}$$