Cartesian product

Printable View

February 26th 2010, 10:15 AM
novice

Cartesian product

My book says the converse of the following implication is false:

$A, B, C, D$ are sets such that $A\subseteq C$ and $B \subseteq D$ , then $A \times B \subseteq C \times D$ .

Although $P \Rightarrow Q \not \equiv Q \Rightarrow P$ , I don't understand why since the above implication still can be true under special circumstances.

For exmple, When $A=D$ and $C=B$ or when $A = B = \emptyset$ or $C=D=\emptyset$ , I still can make it true.

I think $P \Rightarrow Q \not \equiv Q \Rightarrow Q$ is not always true. Am I missing something?
February 26th 2010, 10:20 AM
Drexel28

Quote:

Originally Posted by novice

My book says the converse of the following implication is false:

$A, B, C, D$ are sets such that $A\subseteq C$ and $B \subseteq D$ , then $A \times B \subseteq C \times D$ .

Although $P \Rightarrow Q \not \equiv Q \Rightarrow P$ , I don't understand why since the above implication still can be true under special circumstances.

For exmple, When $A=D$ and $C=B$ or when $A = B = \emptyset$ or $C=D=\emptyset$ , I still can make it true.

I think $P \Rightarrow Q \not \equiv Q \Rightarrow Q$ is not always true. Am I missing something?

Let $(x,y)\in A\times B$ then $x\in A\text{ and }y\in B\implies x\in C\text{ and }y\in D\implies (x,y)\in C\times D$ . Done.
February 26th 2010, 10:59 AM
Plato

Quote:

Originally Posted by novice

My book says the converse of the following implication is false:
$A, B, C, D$ are sets such that $A\subseteq C$ and $B \subseteq D$ , then $A \times B \subseteq C \times D$ .

Consider: $A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\}$ .
February 26th 2010, 11:08 AM
novice

Quote:

Originally Posted by Plato

Consider: $A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\}$ .

You are quite amazing. How could I be so blind?
February 26th 2010, 11:22 AM
novice

Quote:

Originally Posted by Drexel28

Let $(x,y)\in A\times B$ then $x\in A\text{ and }y\in B\implies x\in C\text{ and }y\in D\implies (x,y)\in C\times D$ . Done.

The implication is not a problem, but the converse.

Plato showed the condition for which when it's false. I thought of an empty set on each side of the inclusion, but never thought of having only one emptyset for the leftside while keeping the rightside nonempty.

This is example or over focusing to the point that I loss the entire picture.
February 26th 2010, 11:30 AM
novice

Quote:

Originally Posted by Plato

Consider: $A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\}$ .

Plato,

Now, with the implication, if I declare that $A, B, C$ , and $D$ being nonempty sets and that $A \not = B$ , I think it will be thought to make the converse false.

Do you agree?