# Set Theory , building a set notation

1. May 10, 2013

### reenmachine

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

Build a set notation for $$\bigcup_{i \in N}R × [i , i + 1]$$

3. The attempt at a solution

$\{(x,y) \in R : x \in R \ \ \exists z \in N \ \ z ≤ y ≤ (z+1)\}$

Last time I tried one of these kind of sets I struggled quite a bit , so I'm interested in knowing how much I can handle those from now on.

thanks!

2. May 10, 2013

### reenmachine

Take note that I read the notation as $$\bigcup_{i \in N}(R × [i , i +1])$$

3. May 10, 2013

### ArcanaNoir

This isn't quite right. Can you visualize the elements in the union? Can you describe them with words? When you put $(x,y) \in \mathbb{R}$, you are describing intervals on the real line, is that what you're going for? It looks to me like you should be describing something in 2 dimensions, not 1 dimension.

4. May 10, 2013

### reenmachine

I knew something looked fishy , seems I didn't learn from my mistakes.

Seems to me that the union of all elements of this set will give us the set R.

(Since [i , i+1] with $i \in N$ will always be in R.)

5. May 10, 2013

### ArcanaNoir

Good point, I hadn't thought of it as the union, I was looking at individual elements. But remember, RxR is not the same as R. Is the difference clear?

6. May 10, 2013

### reenmachine

Not sure.

My understanding of it is that $R$ will have elements in the form of $r$ while $R × R$ will have elements in the form of $(r,r)$.But my point was that if you unionize $R$ or $R × R$ , you will end up with the same elements if you ""deconstruct"" the ordered pairs that are made of $r$ in $(r,r)$.

The confusion is probably why should we called them componants? Can a number be a componant as well as an element of a set (the ordered pair as a set)?

7. May 10, 2013

### reenmachine

I will attempt to express what I mean in a clearer way.

Let's take the set $R × [i , i+1]$ with $i \in N$.We accept that all numbers between $i$ and $i+1$ including both will be elements of $R$ as well.$R$ in $R × [i , i+1]$ means that the first member of the ordered pairs will be any real number.So the ordered pair would look like $(x,y)$ with $x$ being any real number and $y$ being any positive real number except maybe all the numbers between $0$ and $1$ if you think $i ≠ 0$ because $0 ∉ N$.Then if you unionize the set $R × [i , i+1]$ , you will something like $\{...(x,y) \cup (x,y) \cup (x,y)...\}$.Since you unionize the ordered pairs , it will give you the set of all elements of those unionized sets(ordered pairs in that case) , therefore will give you all elements of $R$ from all possible $x$ in$(x,y)$.

8. May 11, 2013

### reenmachine

Another attempt :

$\{\{x\},\{x,y\} |\ x,y \in R \ \ y ≥ 1 \}$

EDIT: I finally checked the book's solution , which is $\{(x,y) : x,y \in R \ \ y ≥ 1\}$

I find it weird that the dummy variables would be put between "()" like it's an ordered pair to present the set such as "the set of all (x,y) such that...".Anybody can enlighten me on using my method versus the one in the book concerning the left side of the notation

Last edited: May 11, 2013
9. May 11, 2013

### christoff

They're essentially equivalent, at least insofar as Kuratowski definition of ordered pairs is concerned. In this definition, the cartesian product of two sets $X$ and $Y$, denoted by $X\times Y$ is defined by the set of all ordered pairs $(x,y)=\{\{x\},\{x,y\}\}$, where the object on the left is just a convenient shorthand for the set on the right. See http://en.wikipedia.org/wiki/Cartesian_product for more information.