Results 1 to 6 of 6

Thread: Element in a Set

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    502

    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 $\displaystyle d \notin D$.
    Last edited by novice; Feb 22nd 2010 at 05:40 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    Quote Originally Posted by novice View Post
    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 $\displaystyle 1\in\mathbb{N}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Quote Originally Posted by novice View Post
    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} $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Plato View Post
    $\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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Quote Originally Posted by novice View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Plato View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. element of N
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 12th 2012, 09:12 PM
  2. Orbits of an element and the Stabilizer of the element
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Dec 24th 2011, 05:41 AM
  3. Least Element
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Nov 21st 2010, 12:57 PM
  4. Replies: 3
    Last Post: Mar 23rd 2010, 07:05 PM
  5. element
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Nov 9th 2009, 06:40 AM

Search Tags


/mathhelpforum @mathhelpforum