Understanding Distributive Law Using Arbitrary Union

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Homework Statement

In the textbook I'm reading it tells me that $A \cup \bigcap B = \bigcap \left\{ A \cup X | X \in B \right\}$ for B not equal to ø

Homework Equations

The Attempt at a Solution

I don't understand how this would work, the left side of the equation creates a set where all of the elements of $A$ are included but the right side of the equation takes the arbitrary intersection of $A \cup X$, so wouldn't this mean that some of the elements of $A$ aren't necessarily included in this newly constructed set as the elements of $A$ aren't by necessity also elements of $X$?

How can the left side of the equation be equal to the right?

 

[Ray Vickson]

Homework Statement

In the textbook I'm reading it tells me that $A \cup \bigcap B = \bigcap \left\{ A \cup X | X \in B \right\}$ for B not equal to ø

Homework Equations

The Attempt at a Solution

I don't understand how this would work, the left side of the equation creates a set where all of the elements of $A$ are included but the right side of the equation takes the arbitrary intersection of $A \cup X$, so wouldn't this mean that some of the elements of $A$ aren't necessarily included in this newly constructed set as the elements of $A$ aren't by necessity also elements of $X$?

How can the left side of the equation be equal to the right?

The only way this makes sense is if $B = \{ X_{\gamma}, \gamma \in \Gamma \}$ is a class class of sets, so when you write $X \in B$ you have $X$ itself is a set in the class $B$. In other words, I think it is saying that
$$A \cup \bigcap_{\gamma \in \Gamma} X_{\gamma}<br /> = \bigcap_{\gamma \in \Gamma} A \cup X_{\gamma}$$

 

[HallsofIvy]

Homework Statement

In the textbook I'm reading it tells me that $A \cup \bigcap B = \bigcap \left\{ A \cup X | X \in B \right\}$ for B not equal to ø

Homework Equations

The Attempt at a Solution

I don't understand how this would work, the left side of the equation creates a set where all of the elements of $A$ are included but the right side of the equation takes the arbitrary intersection of $A \cup X$, so wouldn't this mean that some of the elements of $A$ aren't necessarily included in this newly constructed set as the elements of $A$ aren't by necessity also elements of $X$?

No, the union of A with any set, by definition, includes all elements of A. You are confusing the union with the intersection, $A\cap X$.

How can the left side of the equation be equal to the right?