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Subset T S (without middle bar)
 
as opposed to elements S
 
 
Universal set = the set of all objects considered(context specific)
 
 
Given S
 
S
 
 
Set operations 
 
complement S = {| S}
 
 
 
 
 
union T  S =  {x|xor x}
 
 
 
 
 
S= ...
 
 
 
 
 
 
 
intersection
 
 
T S = {|}
 
 
 
Note: this is the first of many pages to be uploaded.  
 
Note: this is the first of many pages to be uploaded.  
 
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[0,1] = {x <math>\in \mathbb{R} </math>such that(s. t.) 0<math>\leq x \leq</math> 1}
 
[0,1] = {x <math>\in \mathbb{R} </math>such that(s. t.) 0<math>\leq x \leq</math> 1}
 
={real positive numbers no greater than 1 as well as 0}
 
={real positive numbers no greater than 1 as well as 0}
 +
 +
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Revision as of 02:48, 15 April 2013

Note: this is the first of many pages to be uploaded.



1/9/13

If S is discrete and finite S = {$ s_1,s_2,s_3 $} S = {head,tail}, S = {win, lose}, S = {1,2,3,4,5,6}

If S is discrete but infinite,

S = {$ s_1,s_2,s_3 $,...} ex. S = {1,2,3,4,...}

    S = {sin(2$ \pi $*440t),sin(2$ \pi $*880t),sin(2$ \pi $*1320t),...}
    Observe $ _{S = \mathbb{R}} $ is not routable; S = [0,1] is not routable
    S = {sin(2$ \pi $*f*t)} f $ \in \mathbb{R} \geq $ 0 
      = {sin(2$ \pi $*f*t)|0$ \leq f < \infty $}

$ \mathbb{Z} $ is all integers $ -\infty $ to $ \infty $

Is $ \mathbb{Z} $ routable? yes.

  $ \mathbb{Z} $={0,1,-1,2,-2,3,-3, }as opposed to $ \mathbb{R} $

$ \mathbb{R} $= {0,3,e,$ \pi $,-1,1.14,$ \sqrt{2} $}

Many different ways to write a set [0,1] = {x $ \in \mathbb{R} $such that(s. t.) 0$ \leq x \leq $ 1} ={real positive numbers no greater than 1 as well as 0}


Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal