# Thread: Element in a Set

1. ## Element in a Set

Let A and B be nonempty sets, and let $\displaystyle (x,y) \in A \times B$.

$\displaystyle (x,y) \in A \times B \Rightarrow x \in A$ and $\displaystyle y \in B$ is logically correct.

Now if C and D are nonempty sets, and that $\displaystyle (c,d) \notin C \times D$.

Will the following be true?

$\displaystyle (x,y) \notin C \times D \Rightarrow c \notin C$ and $\displaystyle d \notin D$.

2. Originally Posted by novice
Let A and B be nonempty sets, and let $\displaystyle (x,y) \in A \times B$.

$\displaystyle (x,y) \in A \times B \Rightarrow x \in A$ and $\displaystyle y \in B$ is logically correct.

Now if C and D are nonempty sets, and that $\displaystyle (c,d) \notin C \times D$.

Will the following be true?

$\displaystyle (x,y) \notin C \times D \Rightarrow c \notin C$ and $\displaystyle d \notin D$.
$\displaystyle (0,1)\notin\mathbb{N}^2$ but $\displaystyle 1\in\mathbb{N}$

3. Originally Posted by novice
Will the following be true?
$\displaystyle (x,y) \notin C \times D \Rightarrow c \notin C$ and $\displaystyle d \notin D$.
$\displaystyle \begin{gathered} \neg \left( {P \wedge Q} \right) \Leftrightarrow \left( {\neg P \vee \neg Q} \right) \hfill \\ \left( {a,b} \right) \notin A \times B \Leftrightarrow a \notin A \vee b \notin B \hfill \\ \end{gathered}$

4. Originally Posted by Plato
$\displaystyle \begin{gathered} \neg \left( {P \wedge Q} \right) \Leftrightarrow \left( {\neg P \vee \neg Q} \right) \hfill \\ \left( {a,b} \right) \notin A \times B \Leftrightarrow a \notin A \vee b \notin B \hfill \\ \end{gathered}$
Plato,
Thank you for pointing the De Morgan's Law. While we are at it, I would like to discuss it with you in regard to a cross product.

Since the De Morgan's Law said $\displaystyle \sim (P \wedge Q) \equiv \sim P \vee \sim Q$.

Now let $\displaystyle A$ and $\displaystyle B$ be nonempty sets and let $\displaystyle (x,y) \in \overline{A \times B}$. We say that $\displaystyle (x,y) \notin A \times B$, which gives us $\displaystyle x \notin A$ or $\displaystyle y \notin B$.

So $\displaystyle x\in \overline{A}$ and $\displaystyle y \in \overline{B}$. Consequently $\displaystyle \overline{A} \times \overline{B}$. We have just shown that $\displaystyle \overline{A \times B} \subset \overline{A} \times \overline{B}$, but according to the De Morgan's Law

$\displaystyle \overline{A \times B} \equiv \overline{A} \times \overline{B}$ is false.

Suppose that I made $\displaystyle \sim (P \wedge Q): \overline{A \times B}$, it should follow that the logic equivalence is $\displaystyle (\sim P \vee \sim Q):$ $\displaystyle \overline{A}$ or $\displaystyle \overline {B}$.

Consequently, $\displaystyle \overline{A}$ or $\displaystyle \overline {B}$ cannot be $\displaystyle \overline{A} \times \overline{B}$.

Do you agree?

5. Originally Posted by novice
Consequently, $\displaystyle \overline{A}$ or $\displaystyle \overline {B}$ cannot be $\displaystyle \overline{A} \times \overline{B}$.
I think that last line is confusing.
With appropriate domain restrictions, this is true: $\displaystyle \overline A \times \overline B \subseteq \overline {A \times B}$.
But they need no be equal.

6. Originally Posted by Plato
I think that last line is confusing.
With appropriate domain restrictions, this is true: $\displaystyle \overline A \times \overline B \subseteq \overline {A \times B}$.
But they need no be equal.
With appropriate domain restrictions----Hmmm.