# Check my proof for cartesian product (set theory)

1. Apr 26, 2010

### iamsmooth

1. The problem statement, all variables and given/known data
Prove that $$\forall$$ sets $$A, B, C$$, if $$B\subseteq C,$$ then $$A \times B \subseteq A \times C$$

2. Relevant equations

3. The attempt at a solution
Haven't done set theory proofs in a while. Does this suffice in proving the statement?:

Let $$x \in B$$ be arbitrary. Assuming $$B \subseteq C$$ is true, we know that $$x \in C$$. Since $$x \in B$$, we know that $$A \times B$$ will produce an ordered pair (a,x) where a is an arbitrary element of A. Since $$x \in C$$, we know that $$A \times C$$ will produce the same ordered pair (a,x).

Therefore by definition of subsets, $$A \times B \subseteq A \times C$$

QED

Last edited: Apr 26, 2010
2. Apr 26, 2010

### tiny-tim

Hi iamsmooth!
mmm … messy …

and what does "$$A \times B$$ will produce an ordered pair (a,x) …" mean?

Instead, start "for any element (a,b) of A x B … "

3. Apr 26, 2010

### iamsmooth

Let $$x \in B$$ be arbitrary. Assuming $$B \subseteq C$$ is true, we know that $$x \in C$$.

We know that for all elements $$(a, x) \in A \times B$$ where a is an arbitrary element of A, there will also exist $$(a, x) \in A \times C$$ since $$x \in A$$ and $$x \in C$$.

Therefore by definition of subsets, $$A \times B \subseteq A \times C$$

QED

Is this better?

4. Apr 26, 2010

### tiny-tim

Since you later say "for all elements $$(a, x) \in A \times B$$", why are you starting with "Let $$x \in B$$ be arbitrary." ?

The proposition you are required to prove starts "for any element (a,x) of A x B",

5. Apr 26, 2010

### iamsmooth

Assuming $$B \subseteq C$$ is true, we know that if there is any element $$x \in B$$, then $$x \in C$$.

For any element $$(a, b) \in A \times B$$ where a is an arbitrary element of A and b is an arbitrary element of B, there will also exist $$(a, b) \in A \times C$$ since $$b \in B$$ and $$b \in C$$ must be true (by assumption).

Therefore by definition of subsets, $$A \times B \subseteq A \times C$$

QED

Is it just my wording that's messed up?

6. Apr 26, 2010

### tiny-tim

You really don't need that opening sentence;

also, "there will also exist" is rather a strange way of putting it, since it's the same element, (a,b), in both sets …

you want something more like any element of A x B is of the form (a,b) with a in A and b in B, and since B is a subset of C, b is in C, and so (a,b) is in A x C.