Sets definiton

1. Jul 4, 2010

annoymage

1. The problem statement, all variables and given/known data

1.Given a set T we say that T serves as an index set for family F={Aa} of sets if for every a in T there exists a set Aa in family F.

2. By the union of the sets Aa, where a is in T, we mean the set
{x l x$$\in$$Aa for at least one a in T}. We shall denote it by $$\bigcup$$(a$$\in$$T) Aa.

i think its better if i show example

example:

if S is the set of real number, T is the set of rational number, let, for

a$$\in$$T, Aa = {x$$\in$$S l x$$\geq$$a}

so $$\bigcup$$(a$$\in$$T) Aa = S

what i dont understand

i can see how ..,A-1, A0, A1 ,... is,

and i dont know how to change it to
$$\bigcup$$(a$$\in$$T) Aa
or how it is equal to S, because im perplexed with definition (2) particularly x$$\in$$Aa for at least one a in T

help help

p/s: sorry if it is abit messy, still progressing in latex

2. Jul 4, 2010

HallsofIvy

Doesn't the definition say "for every a in T there exists exactly one set Aa in family F"?

Do you mean what those sets are?
A-1 is, by this definition, the set of all real numbers greater than or equal to 0: $\{x| x\ge -1\}= [-1, \infty)$. $A_0= [0, \infty)$, etc.

What do you mean "change" it to that? They are not the same at all- one is a collection of sets, the other is the union of all those- the set of all numbers in any one of them.

That simply means "x is in at least one of those sets". The union of a collection of sets is, as usual, the set of all members of any of the sets in the collection.

In this particular case, since, given any real number x, there exist a rational number, r< x, $x\in A_r$ for that particular r. Since every real number is in at least one of those sets, S is just the set of all real numbers.

Last edited by a moderator: Jul 5, 2010
3. Jul 4, 2010

weaselman

It's quite simple really, just a lot of fancy notation.
So, you got a bunch of As, each contains a bunch of x's. Get all the unique x's, from all the A's and put them together into a new set. This new set is S. It contains all the x's such that each x belongs it some of the A's.

I hope it helps.

4. Jul 4, 2010

annoymage

no, i copy this from "Topic in Algebra, Herstien",
but yea, the "there exist a set" confuses me, like, for every a , there's maybe other branches of set.

yes, yes, thats what i mean, i'm still progressing in english, so, correct me if i'm wrong along the way.

my mistake

ok, so i confused here,

A(-1)=[-1,oo), A(0)=[0,oo], A(1)=[1,oo),.. these are the sets,

"Since every real number is in at least one of those sets"

so, let 1 for example, A(2) dont have {1}?