Check my proof for cartesian product (set theory)

[iamsmooth]

Homework Statement

Prove that \forall sets A, B, C, if B\subseteq C, then A \times B \subseteq A \times C

Homework Equations

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

QEDThanks for your help.

 

[tiny-tim]

Hi iamsmooth!

Prove that \forall sets A, B, C, if B\subseteq C, then A \times B \subseteq A \times C
…
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

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 … "

 

[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?

 

[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",

so your proof should start with the same words.

 

[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?

 

[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.