1. ## 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

2. 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)"

3. Originally Posted by powderteacher
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

4. ## 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
% aaqaaiabeU7aSjabgIGiolabfU5ambqab0GaeSOkIufaaOGaay jkai
% BaaaleaacqaH7oaBaeqaaOGaaiykaaWcbaGaeq4UdWMaeyicI4 Saeu
% baGaamyqamaaBaaaleaacqaH7oaBaeqaaaqaaiabeU7aSjabgI Giol
% abfU5ambqab0GaeSykIKeaaOGaayjkaiaawMcaaiabg2da9maa taba
% GaaiykaaWcbaGaeq4UdWMaeyicI4Saeu4MdWeabeqdcqWIQisv aaaa
% aa!67B5!
$\begin{gathered} S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\ S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\ \end{gathered}$

5. ## 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} 6. 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 $$and$$ tags quote my post to see what i posted, or just click on the code, either way see here 7. 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)}

$? 8. You all have wasted a good deal of effort: because I answered that question in the first response. Why don’t you read before responding? 9. Originally Posted by powderteacher I got it thank you very much DeMorgan's$\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 \}$which is what we had before...in one step 10. Originally Posted by Jhevon as i said, DeMorgan's law is overkill here. i would not waste time applying DeMorgan's law. but, by DeMorgan's law$\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