Cartesian product

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• 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?
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