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Thread: DeMorgan's laws help

  1. #1
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    DeMorgan's laws help

    Hello,
    I am currently making my way through the preliminaries of my Intro to Analysis book and I need clarification on a few things. I am confused on how we can used the indexed family of sets notation to describe DeMorgan's Laws. How can I write with mathtype on these posts? Anyways my question is how can i use demorgans laws to write a simpler description of:
    R\(the intersection from n=1 to inf of the indexed family of sets defined by
    (-1/n,1/n)? Any help would be much appreciated, thanks
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  2. #2
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    For me it is diffcult to know what you mean.
    But $\displaystyle R\backslash \bigcap\limits_{n = 1}^\infty {\left( {\frac{{ - 1}}{n},\frac{1}{n}} \right)} = \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$
    is coded by "R\backslash \bigcap\limits_{n = 1}^\infty {\left( {\frac{{ - 1}}{n},\frac{1}{n}} \right)} = \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)"
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by powderteacher View Post
    Hello,
    I am currently making my way through the preliminaries of my Intro to Analysis book and I need clarification on a few things. I am confused on how we can used the indexed family of sets notation to describe DeMorgan's Laws. How can I write with mathtype on these posts? Anyways my question is how can i use demorgans laws to write a simpler description of:
    R\(the intersection from n=1 to inf of the indexed family of sets defined by
    (-1/n,1/n)? Any help would be much appreciated, thanks
    what is $\displaystyle \bigcap_{n = 1}^{\infty} \left( - \frac 1n, \frac 1n \right)$ ?

    by DeMrogan's Laws, i assume you are referring to:

    $\displaystyle C - (A \cap B) = (C - A) \cup (C - B)$

    $\displaystyle C - (A \cup B) = (C - A) \cap (C - B)$

    i don't see how DeMorgan's laws make this simpler. i'd just write it as $\displaystyle \mathbb{R} \backslash \{ 0 \} = (-\infty,0) \cup (0, \infty)$ or something...i gave away the answer

    we use LaTex here. you can paste codes from MathType, but you have to change the tags
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  4. #4
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    Smile thanks

    I'm not sure how Demorgans makes this any simpler either. Anyways thank you for the latex help, and the solution.
    I have DeMorgan's as:
    % MathType!MTEF!2!1!+-
    % feaaeaart1ev0aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL wBLn
    % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt ubsr
    % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb
    % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr
    % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaam4u ai
    % aacYfadaqadaqaamaatababaGaamyqamaaBaaaleaacqaH7oaB aeqa
    % aaqaaiabeU7aSjabgIGiolabfU5ambqab0GaeSOkIufaaOGaay jkai
    % aawMcaaiabg2da9maauababaGaaiikaiaadofacaGGCbGaamyq amaa
    % BaaaleaacqaH7oaBaeqaaOGaaiykaaWcbaGaeq4UdWMaeyicI4 Saeu
    % 4MdWeabeqdcqWIPissaaGcbaGaam4uaiaacYfadaqadaqaamaa uaba
    % baGaamyqamaaBaaaleaacqaH7oaBaeqaaaqaaiabeU7aSjabgI Giol
    % abfU5ambqab0GaeSykIKeaaOGaayjkaiaawMcaaiabg2da9maa taba
    % baGaaiikaiaadofacaGGCbGaamyqamaaBaaaleaacqaH7oaBae qaaO
    % GaaiykaaWcbaGaeq4UdWMaeyicI4Saeu4MdWeabeqdcqWIQisv aaaa
    % aa!67B5!
    \[
    \begin{gathered}
    S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\
    S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\
    \end{gathered}
    \]
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  5. #5
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    oops lets try that again

    DeMorgan's Laws:
    $\displaystyle % MathType!MTEF!2!1!+-
    % feaaeaart1ev0aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL wBLn
    % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt ubsr
    % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb
    % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr
    % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaam4u ai
    % aacYfadaqadaqaamaatababaGaamyqamaaBaaaleaacqaH7oaB aeqa
    % aaqaaiabeU7aSjabgIGiolabfU5ambqab0GaeSOkIufaaOGaay jkai
    % aawMcaaiabg2da9maauababaGaaiikaiaadofacaGGCbGaamyq amaa
    % BaaaleaacqaH7oaBaeqaaOGaaiykaaWcbaGaeq4UdWMaeyicI4 Saeu
    % 4MdWeabeqdcqWIPissaaGcbaGaam4uaiaacYfadaqadaqaamaa uaba
    % baGaamyqamaaBaaaleaacqaH7oaBaeqaaaqaaiabeU7aSjabgI Giol
    % abfU5ambqab0GaeSykIKeaaOGaayjkaiaawMcaaiabg2da9maa taba
    % baGaaiikaiaadofacaGGCbGaamyqamaaBaaaleaacqaH7oaBae qaaO
    % GaaiykaaWcbaGaeq4UdWMaeyicI4Saeu4MdWeabeqdcqWIQisv aaaa
    % aa!67B5!
    \[
    \begin{gathered}
    S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\
    S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\
    \end{gathered}

    $
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    i think you meant this:
    $\displaystyle
    \begin{gathered}
    S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\
    S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\
    \end{gathered}
    $

    as i said, you have to change the \[ and \] tags to [tex] and [/tex] tags

    quote my post to see what i posted, or just click on the code, either way

    see here
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  7. #7
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    I got it thank you very much
    DeMorgan's
    $\displaystyle
    S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )}

    $

    $\displaystyle
    S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )}

    $

    But how can I use the second one to simplify:
    $\displaystyle
    R\backslash \bigcap\limits_{n = 1}^\infty {\left( { - \frac{1}
    {n},\frac{1}
    {n}} \right)}

    $ ?
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  8. #8
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    You all have wasted a good deal of effort: because I answered that question in the first response.
    Why donít you read before responding?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by powderteacher View Post
    I got it thank you very much
    DeMorgan's
    $\displaystyle
    S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )}

    $

    $\displaystyle
    S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )}

    $

    But how can I use the second one to simplify:
    $\displaystyle
    R\backslash \bigcap\limits_{n = 1}^\infty {\left( { - \frac{1}
    {n},\frac{1}
    {n}} \right)}

    $ ?
    as i said, DeMorgan's law is overkill here. i would not waste time applying DeMorgan's law.

    but, by DeMorgan's law $\displaystyle \mathbb{R} \backslash \bigcap_{n = 1}^{\infty} \left( - \frac 1n, \frac 1n \right) = \bigcup_{n = 1}^{\infty} \left[ \mathbb{R} \backslash \left( - \frac 1n, \frac 1n \right) \right] = $ $\displaystyle \mathbb{R} \backslash (-1,1) \cup \mathbb{R} \backslash (-1/2, 1/2) \cup \mathbb{R} \backslash (-1/3,1/3) \cdots = \mathbb{R} \backslash \{ 0 \}$ which is what we had before...in one step
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  10. #10
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    Smile

    Quote Originally Posted by Jhevon View Post
    as i said, DeMorgan's law is overkill here. i would not waste time applying DeMorgan's law.

    but, by DeMorgan's law $\displaystyle \mathbb{R} \backslash \bigcap_{n = 1}^{\infty} \left( - \frac 1n, \frac 1n \right) = \bigcup_{n = 1}^{\infty} \left[ \mathbb{R} \backslash \left( - \frac 1n, \frac 1n \right) \right] = $ $\displaystyle \mathbb{R} \backslash (-1,1) \cup \mathbb{R} \backslash (-1/2, 1/2) \cup \mathbb{R} \backslash (-1/3,1/3) \cdots = \mathbb{R} \backslash \{ 0 \}$ which is what we had before...in one step
    thank you that is exactly what i needed help with
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