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?