Let A, B, and C be sets with A not equal to the empty set...

**Brjakewa** · November 13th 2011, 10:23 AM

Is the following proposition true or false? Justify your conclusion.

Let A, B, and C be sets with A not equal to the empty set. If the Cartesian Product AXB = the Cartesian Product AXC, then B=C.

Explain where the assumption that A is not equal to the empty set is needed.

This proposition seems true to me, but I have no idea how to prove it. I tried showing that AXB = AXC by showing they are subsets of each other, but I'm not sure if that's the right approach. Additionally, my proof of showing they are subsets of one another didn't get very far.

Can anyone show me how to do this step by step?

**veileen** · November 13th 2011, 10:36 AM

$A\times B=A\times C$ , so the cardinal of $A\times B$ is equal to the cardinal of $A\times C$ , so the cardinal of B is equal to the cardinal of C.

If $B = \left \{ b_1, b_2, ..., b_n \right \}$ and $C = \left \{ c_1, c_2, ..., c_n \right \}$ , write $A\times B$ and $A\times C$ , then... I don't know it's pretty obvious. ^^'

**Plato** · November 13th 2011, 10:43 AM

Originally Posted by Brjakewa

Is the following proposition true or false? Justify your conclusion.
Let A, B, and C be sets with A not equal to the empty set. If the Cartesian Product AXB = the Cartesian Product AXC, then B=C.

Explain where the assumption that A is not equal to the empty set is needed.

We know that $\left( {\exists a \in A} \right)$ , WHY?

If $x\in B$ then $(a,x)\in A\times B.$ Why?

Does that mean $(a,x)\in A\times C~?$ WHY?

So does that mean $x\in C~?$ WHY?

YOU finish it now.

**Plato** · November 13th 2011, 10:48 AM

Originally Posted by veileen

$A\times B=A\times C$ , so the cardinal of $A\times B$ is equal to the cardinal of $A\times C$ , so the cardinal of B is equal to the cardinal of C.

If $B = \left \{ b_1, b_2, ..., b_n \right \}$ and $B = \left \{ c_1, c_2, ..., c_n \right \}$ , write $A\times B$ and $A\times C$ , then... I don't know it's pretty obvious. ^^'

How would that even begin to show $B=C~?$
This proof is not about cardinalty.

**veileen** · November 13th 2011, 11:09 AM

The second set is $C = \left \{ c_1, c_2, ..., c_n \right \}$ , sorry.

$A \times B=A \times C$ means that every element from $A \times B$ is in $A \times C$ too.

**Plato** · November 13th 2011, 11:41 AM

Originally Posted by veileen

The second set is $C = \left \{ c_1, c_2, ..., c_n \right \}$ , sorry.

Why are you assuming that sets $C~\&~B$ are finite?
Again, cardinality has nothing to do with this problem.

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