# [SOLVED] Finding the union and intersection of indexed sets.

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• Oct 13th 2008, 08:37 PM
ilikedmath
[SOLVED] Finding the union and intersection of indexed sets.
I did not want to start 6 different threads so I am posting them all here since they all refer to the same problem which is:

Find the union and intersection of the collection of sets.

Any help on any of the problems would be great. I think I've gotten an idea for some of them. I'm stumped on 2 of the 6, and have some work for the rest.

(N is the set of natural numbers)
Sets have the form: (-1/n , 1) meaning all real numbers in that interval
http://www.cramster.com/Answer-Board...0837502331.gif= (-1, 1) because -1/n gets smaller as n gets larger so -1/n gets closer to 0. So the largest of sets will contain all other sets so that's how I got the union to be (-1, 1).
http://www.cramster.com/Answer-Board...7712507077.gif= [0, 1) because 0 will be in all sets and it will never reach 1 because of the definition of An

(2) http://www.cramster.com/Answer-Board...0212507093.gif (Q is the set of rational numbers)
Sets have the form [x, x + 1] meaning all rationals in that interval
http://www.cramster.com/Answer-Board...6927500483.gif= (-∞, ∞) because I can put any real number in for x and so the union would have to contain all of the sets and since the reals go on and on both ways I saw (-∞, ∞) as the union.

(3) http://www.cramster.com/Answer-Board...7712502182.gif (Q is the set of rational numbers)
Sets have the form: [r , r + 1] meaning all real numbers in that interval
-- this is one the questions that stumped me. I am having a hard time seeing the union and intersection. This collection of sets does look very similar to the previous problem but it switches the use of real and rational numbers. I tried to list out some sets (such as E0, E-1/2, E3/5, E3) to get some ideas, but I still couldn't come to some answer.

To get an idea, I wrote some sets out such as E.00001, E.00052, E.123, and E.9999.
So the end points of the intervals get closer and closer to -1 and 1 but never equal -1 and 1 because x is in (0, 1). So I got that:
http://www.cramster.com/Answer-Board...2087508860.gif= (-1,1) and http://www.cramster.com/Answer-Board...3962508034.gif= {0} since 0 is in all intervals of the form [(the negative of a number), (the number)].

(5) http://www.cramster.com/Answer-Board...2865000787.gif (N is the set of naturals)
Sets have the form: [0, 1/n) meaning all irrationals in that interval
The right side of the interval is getting smaller as n increases so [0, 1) is the largest which will contain all other smaller sets so I got:
http://www.cramster.com/Answer-Board...5525009491.gif= [0,1) and http://www.cramster.com/Answer-Board...6150004968.gif= {0} because 0 is in all the sets by the form [0, 1/n).

(6) http://www.cramster.com/Answer-Board...9587504353.gif (R\Q is the set of all irrational numbers)
This is the other question that stumped me. I could not figure out the form the sets would be in. Is the form "z < n"? And where is this n coming from?
I tried to list some sets but couldn't see a pattern nor get an idea towards an answer.
For example http://www.cramster.com/Answer-Board...0212508992.gif= http://www.cramster.com/Answer-Board...2712502201.gif so that means it is the set of all naturals greater than -√3 ?
I am quite confused with this one.

Any help in any form (hints, tips, suggestions, corrections, etc.) is greatly appreciated! :)
• Oct 14th 2008, 03:39 AM
Plato
In #2 there are no irrational numbers in any of the set.
So how could the union be all real numbers?

In #4, $\displaystyle E_{1/10} = \left[ { - 10,10} \right]\;\& \;E_{1/100} = \left[ { - 100,100} \right]$.
So you best rethink all of these.

In is best to post at most two questions at a time.
You are more likely to get real help.
• Oct 14th 2008, 03:50 AM
vincisonfire
For number 2 the answer is good but remember that this is true in the set of rational. That is the union contains all rational numbers (-infinity , infinity) in Q.
• Oct 14th 2008, 03:55 AM
vincisonfire
For number 3, consider the following rational numbers : ... -2, -1, 0, 1, 2 ...
These are the integers. If you take the union on set of real from 0 to 1 and 1 to 2 etc you'll get the set of real. If you had the other intervals that are rational bounded eg 1/2 to 3/2 you simply cover part of real line that are already done.
From this, it is easy to see that the intersection is the empty set.
• Oct 14th 2008, 03:58 AM
vincisonfire
Plato is right for question 4. Answer would be more likely to be (-infinity , infinity). The best way to answer these questions are to plug some values in the definition to see what the sets look like. Answers are often obvious once you've done this.
• Oct 14th 2008, 04:03 AM
vincisonfire
The notation R\Q means all real numbers EXCEPT rational. You thus know that z in never rational.
You know that for set An, z is smaller than n that is a natural number.
Eg A10 contains all real number smaller than 10 that are not rational.
I would say that union contains R\Q and the intersection should be the empty set.
• Oct 14th 2008, 05:32 AM
ilikedmath
Quote:

Originally Posted by Plato
In #2 there are no irrational numbers in any of the set.
So how could the union be all real numbers?

In #4, $\displaystyle E_{1/10} = \left[ { - 10,10} \right]\;\& \;E_{1/100} = \left[ { - 100,100} \right]$.
So you best rethink all of these.

In is best to post at most two questions at a time.
You are more likely to get real help.

Oops, yeah that's right. I see how the union can't be all reals for #2.

Hmm...so for #4, would the union be all reals then if when I plug in decimals such as those you mentioned, I end up with their reciprocals? Is the intersection correct that it is {0}?

So if I repost this breaking it up into 2 questions per post, can I have someone delete this longer post so it's not repeated?
• Oct 14th 2008, 05:38 AM
Plato
Here is a strong hint on the intersection in #4.
$\displaystyle \left( {\forall x \in (0,1)} \right)\left[ { - \frac{1} {x} < - 1 < 1 < \frac{1} {x}} \right]$
• Oct 14th 2008, 05:39 AM
ilikedmath
Quote:

Originally Posted by vincisonfire
The notation R\Q means all real numbers EXCEPT rational. You thus know that z in never rational.
You know that for set An, z is smaller than n that is a natural number.
Eg A10 contains all real number smaller than 10 that are not rational.
I would say that union contains R\Q and the intersection should be the empty set.

Thanks for all your help! I'll rework these and also repost these questions separately as Plato suggested, so I can get more/other feedback as well. I'd really like to understand these better so the more help I get the better. Thanks again!
• Oct 14th 2008, 05:41 AM
ilikedmath
Quote:

Originally Posted by Plato
Here is a strong hint on the intersection in #4.
$\displaystyle \left( {\forall x \in (0,1)} \right)\left[ { - \frac{1} {x} < - 1 < 1 < \frac{1} {x}} \right]$

How do I know -1 and 1 are in the interval of [-1/x, 1/x]? Is it because of those certain values you'd mentioned such as 1/10, 1/100, etc.?
• Oct 14th 2008, 05:45 AM
Plato
Quote:

Originally Posted by ilikedmath
How do I know -1 and 1 are in the interval of [-1/x, 1/x]? Is it because of those certain values you'd mentioned such as 1/10, 1/100, etc.?

Is this a true statement: $\displaystyle \left( {\forall x \in (0,1)} \right)\left[ { - \frac{1}{x} < - 1 < 1 < \frac{1}{x}} \right]$?
Now that is for every x in (0,1).
• Oct 14th 2008, 06:00 AM
ilikedmath
Quote:

Originally Posted by Plato
Is this a true statement: $\displaystyle \left( {\forall x \in (0,1)} \right)\left[ { - \frac{1}{x} < - 1 < 1 < \frac{1}{x}} \right]$?
Now that is for every x in (0,1).

How do I know it's true?
Thanks for all your help thus far.
I'll look into these more later after my class. I will post again once I've come up with more logic behind my answers. It does me no good to have answers and not be able to explain how/why I got them. Thanks for your time!
• Oct 14th 2008, 06:09 AM
Plato
Quote:

Originally Posted by ilikedmath
How do I know it's true?

$\displaystyle \begin{gathered} \left( {\forall x \in (0,1)} \right)\left[ {0 < x < 1} \right] \hfill \\ 0 < x < 1 \Rightarrow \quad \frac{1} {x} > 1 \Rightarrow \quad - \frac{1} {x} < - 1 \hfill \\ \end{gathered}$
• Oct 14th 2008, 08:51 AM
ilikedmath
Quote:

Originally Posted by Plato
$\displaystyle \begin{gathered} \left( {\forall x \in (0,1)} \right)\left[ {0 < x < 1} \right] \hfill \\ 0 < x < 1 \Rightarrow \quad \frac{1} {x} > 1 \Rightarrow \quad - \frac{1} {x} < - 1 \hfill \\ \end{gathered}$

(Rofl) <<< that is me laughing at myself for missing such a simple algebraic reasoning! Thanks for clarifying that!
• Oct 14th 2008, 07:10 PM
ilikedmath
Quote:

Originally Posted by vincisonfire
For number 2 the answer is good but remember that this is true in the set of rational. That is the union contains all rational numbers (-infinity , infinity) in Q.

I think I put all real numbers because of the "$\displaystyle \leq$" sign so that means the end points are included in the sets, right? So for every set, it has all rationals in the interval, but what if the end points are irrational numbers? They're still in the set aren't they? So I think that's why I put all reals. If I am seeing/thinking about this wrong, please let me know. This is due tomorrow :|
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