# I Intersection & union of closed and open sets?

1. Sep 7, 2016

### arupel

I am a little confused here:

a) The number 2 which is at the beginng of one set is closed.
The number 2 is open at the beginning of the other set.

b) The number 2 is closed of the beginning of a set which goes to infinity.
The other set begins at 0 and goes to infinity (2 is an element of the set).

How is closed 2 handled in the interaction and union of the sets in a and b?
Another way of asking these questions is how do the line intervals look like?

Thanks,
Confused,
Arthur Rupel

2. Sep 7, 2016

### arupel

Excuse me. I said interaction. I meant intersection.

Art Rupel

3. Sep 7, 2016

### Staff: Mentor

Could you specify the sets and where you take them from? And 2 as well. Are we talking about real numbers?

4. Sep 7, 2016

### Staff: Mentor

I'm going to assume that the sets you're asking about are intervals on the real number line.
The first set above is the interval $[2, \infty)$, which can also be written as the inequality $x \ge 2$.
The second set is the interval $(2, \infty)$. As an inequality, it would be x > 2.

We don't say that a number is closed or open; rather the set is closed or open (in addition to other possibilities). An endpoint of an interval can be included or not included.
First set above is the same as the first one in part a) above.
Second set: as an interval it could be $(0, \infty)$ or $[0, \infty)$. You didn't specify whether the left endpoint (0) was included or not. An inequality that represents the first interval is $x > 0$. For the second interval, $x \ge 0$.
The intersection of two sets consists of all those elements that belong to both sets.
The union of two sets consists of all those elements that are in the first set or in the second set or in both sets, if the sets intersect.

Again, "closed 2" doesn't make any sense. If 2 is an element of one set (or one interval), it will also be in the union of that set and whatever other set you're working with. If 2 is an element of both sets, it will be in the intersection of those two sets.

5. Sep 7, 2016

### arupel

I am using 2 as an example. I was t thinking of integers, but an answer in real numbers would also be interesting.
These are just examples for heuristic purposes.

In retospection (as is alway true), I think I answered b). In union and intersection 2 is just an element in the intersection and union of both sets.

But, (also in retrospection) another case:
2 is open at the beginning of a set which goes to infinity.
The other set begins at zero and goes to infinity.

Im guessing that the intersection of both sets (for real numbers), since 2 is open (approached but never an element in one set) the intersection of both sets cannot include 2. You can get as close to 2 (for real numbers) but you want, but you can never touch it .

For intergers, the next integer is 3 which is a element of the intersection of both sets.

In the union of both sets 2 is element, since it is in one of the sets.

Sorry for this messiness, but I keeping thinking long after I write.

Again,
Thanks
Art Rupel

6. Sep 7, 2016

### Staff: Mentor

I really don't think the sets contain only integers, based on your use of "open" and "closed." As far as I can tell, your sets are intervals, sets that include all of the real numbers between two specified numbers, or all numbers from one specified number to infinity.
Yes, and so are an infinite number of other numbers.
If you want to be understood, don't say "2 is open." The interval is open.

This set can be described in interval notation as $(2, \infty)$.
The the left parenthesis indicates that 2 is not included in the interval. If the interval extends infinitely far, we always use a right parenthesis following the infinity symbol.
The interval can also be described by an inequality: x > 2. In set builder notation, this is {x | x > 2}; i.e., "the set of numbers x such that x is greater than 2."

This is a lot of words that don't add up to a description of the intersection of the two sets.
Again, we don't talk about numbers being open or closed. They either are or aren't included as elements of some set.

If $A = (2, \infty)$ and B = $[0, \infty)$, then
$A \cup B = [0, \infty)$ (the union of A and B), and
$A \cap B = (2, \infty)$ (the intersection of A and B)
It turns out here that $A \cup B$ is the same as B (i.e., equal to B), which includes 0, 0.7, 1.3, 2, 2.01, and an infinity of other numbers. In short any number greater than or equal to 0.

And $A \cap B$ turns out to be the same as A. The set does not include 2, but it doesn include 2.1, 2.01, 2.001, and numbers that are arbitrarily close to 2, but still larger than 2.

7. Sep 8, 2016

### arupel

Thanks, I am getting my feet wet on closed and open intervals. Just one more question:

A = (2, infinity)
B = [0, infinity)

Intersection of A & B is (2,infinity)?
Union of A & B is [0,infinity)?
Thanks,
Art Rupel

8. Sep 8, 2016

### Staff: Mentor

Yes, which is what I said in post #6.