Math Help - DeMorgan's laws help

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 $R\backslash \bigcap\limits_{n = 1}^\infty {\left( {\frac{{ - 1}}{n},\frac{1}{n}} \right)} = \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$
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 $\bigcap_{n = 1}^{\infty} \left( - \frac 1n, \frac 1n \right)$ ?

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

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

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

i don't see how DeMorgan's laws make this simpler. i'd just write it as $\mathbb{R} \backslash \{ 0 \} = (-\infty,0) \cup (0, \infty)$ 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
% 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaam4u ai
% aacYfadaqadaqaamaatababaGaamyqamaaBaaaleaacqaH7oaB aeqa
% aaqaaiabeU7aSjabgIGiolabfU5ambqab0GaeSOkIufaaOGaay jkai
% aawMcaaiabg2da9maauababaGaaiikaiaadofacaGGCbGaamyq amaa
% BaaaleaacqaH7oaBaeqaaOGaaiykaaWcbaGaeq4UdWMaeyicI4 Saeu
% 4MdWeabeqdcqWIPissaaGcbaGaam4uaiaacYfadaqadaqaamaa uaba
% baGaamyqamaaBaaaleaacqaH7oaBaeqaaaqaaiabeU7aSjabgI Giol
% abfU5ambqab0GaeSykIKeaaOGaayjkaiaawMcaaiabg2da9maa taba
% baGaaiikaiaadofacaGGCbGaamyqamaaBaaaleaacqaH7oaBae qaaO
% GaaiykaaWcbaGaeq4UdWMaeyicI4Saeu4MdWeabeqdcqWIQisv aaaa
% aa!67B5!
$\begin{gathered} S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\ S\backslash \left( {\bigcap\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcup\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )} \hfill \\ \end{gathered}$

5. oops lets try that again

DeMorgan's Laws:
$% 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: $ \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
$
S\backslash \left( {\bigcup\nolimits_{\lambda \in \Lambda } {A_\lambda } } \right) = \bigcap\nolimits_{\lambda \in \Lambda } {(S\backslash A_\lambda )}

$

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

$

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