Proving Equality of Sets in Algebra

[jonroberts74]

Homework Statement

Prove

[tex]A \times (B \cap C) = (A \times B) \cap (A \times C) [/tex]

The Attempt at a Solution

Let [tex]x \in A[/tex] and [tex]y \in B \cap C \rightarrow y \in B \wedge y \in C[/tex]

now [tex] \exists (x,y) \in A \times (B \cap C) [/tex]

so [tex](x,y) \in A \times B \wedge (x,y) \in A \times C[/tex]

thus [tex](x,y) \in (A \times B) \cap (A \times C) [/tex]

therefore [tex]A \times (B \cap C) = (A \times B) \cap (A \times C) [/tex]

 

[verty]

A,B,C could all be the empty set, in which case your third line is incorrect, there may not exist such (x,y).

I would use set-builder notion to show that they are the same.

 

[jonroberts74]

If they are all the empty set then that's pretty trivial and not interesting. And all the same? As in A =B=C?

 

[verty]

If they are all the empty set then that's pretty trivial and not interesting. And all the same? As in A =B=C?

I mean the left and right-hand sides, you want to show that they are the same set.

 

[SammyS]

Homework Statement

Prove [tex]A \times (B \cap C) = (A \times B) \cap (A \times C) [/tex]

The Attempt at a Solution

Let [tex]x \in A[/tex] and [tex]y \in B \cap C \rightarrow y \in B \wedge y \in C[/tex]
now [tex] \exists (x,y) \in A \times (B \cap C) [/tex]
so [tex](x,y) \in A \times B \wedge (x,y) \in A \times C[/tex]
thus [tex](x,y) \in (A \times B) \cap (A \times C) [/tex]
therefore, [tex]A \times (B \cap C) = (A \times B) \cap (A \times C) [/tex]

I addition to what verty pointed out:

You have only done half of the proof.

You showed that the left hand side is a subset of the right hand side.

 

[jonroberts74]

I mean the left and right-hand sides, you want to show that they are the same set.

Ah yes. I went back and showed it goes both ways. Thanks