Proving Equality of Sets in Algebra

[jonroberts74]

Homework Statement

Prove

$$A \times (B \cap C) = (A \times B) \cap (A \times C)$$

The Attempt at a Solution

Let $$x \in A$$ and $$y \in B \cap C \rightarrow y \in B \wedge y \in C$$

now $$\exists (x,y) \in A \times (B \cap C)$$

so $$(x,y) \in A \times B \wedge (x,y) \in A \times C$$

thus $$(x,y) \in (A \times B) \cap (A \times C)$$

therefore $$A \times (B \cap C) = (A \times B) \cap (A \times C)$$

 

[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 $$A \times (B \cap C) = (A \times B) \cap (A \times C)$$

The Attempt at a Solution

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

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