# Element in a Set

• February 22nd 2010, 05:17 PM
novice
Element in a Set
Let A and B be nonempty sets, and let $(x,y) \in A \times B$.

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

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

Will the following be true?

$(x,y) \notin C \times D \Rightarrow c \notin C$ and $d \notin D$.
• February 22nd 2010, 06:20 PM
Drexel28
Quote:

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

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

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

Will the following be true?

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

$(0,1)\notin\mathbb{N}^2$ but $1\in\mathbb{N}$
• February 23rd 2010, 06:55 AM
Plato
Quote:

Originally Posted by novice
Will the following be true?
$(x,y) \notin C \times D \Rightarrow c \notin C$ and $d \notin D$.

$\begin{gathered}
\neg \left( {P \wedge Q} \right) \Leftrightarrow \left( {\neg P \vee \neg Q} \right) \hfill \\
\left( {a,b} \right) \notin A \times B \Leftrightarrow a \notin A \vee b \notin B \hfill \\ \end{gathered}$
• February 23rd 2010, 08:26 AM
novice
Quote:

Originally Posted by Plato
$\begin{gathered}
\neg \left( {P \wedge Q} \right) \Leftrightarrow \left( {\neg P \vee \neg Q} \right) \hfill \\
\left( {a,b} \right) \notin A \times B \Leftrightarrow a \notin A \vee b \notin B \hfill \\ \end{gathered}$

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 $\sim (P \wedge Q) \equiv \sim P \vee \sim Q$.

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

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

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

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

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

Do you agree?
• February 23rd 2010, 08:51 AM
Plato
Quote:

Originally Posted by novice
Consequently, $\overline{A}$ or $\overline {B}$ cannot be $\overline{A} \times \overline{B}$.

I think that last line is confusing.
With appropriate domain restrictions, this is true: $\overline A \times \overline B \subseteq \overline {A \times B}$.
But they need no be equal.
• February 23rd 2010, 09:07 AM
novice
Quote:

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

With appropriate domain restrictions----(Thinking)Hmmm.