# Find a formula for the intersection math

1. Sep 8, 2006

### ZioX

Let I=(a,b) and J=(c,d) with I and J having a nonempty intersection. Find a formula for the intersection of I and J and prove it.

When a<c I have found that the intersection is [c,b]. Now I need to prove that it is. The way I intend to prove it is by showing that [c,b] and the intersection of I and J are both subsets of each other. I've done one, [c,b] being a subset of I intersect J.

I intersect J being a subset of [c,b] is a bit trickier. I've broken it down to:

let x be an element of I intersect J then x is an element of I and x is an element of J. Then:

a<=x<=b and c<=x<=d.

Now this is where I'm stumped.

2. Sep 8, 2006

### JasonRox

Are I and J sets? Like on the real line?

3. Sep 8, 2006

### JasonRox

If a<c, how do you know if c<b to make an intersection of [c,b]? And, how can an intersection of two open sets become closed?

If a<c and b<c, then there is no intersection.

4. Sep 8, 2006

### ZioX

Ooops. Yes, I and J are intervals. Intersection is nonempty.

5. Sep 8, 2006

### JasonRox

There's a formula for solving this?

Note: The intersection of two open intervals is an open interval, so I don't know why you represent the intersection as a closed interval.

You said if a<c, then the intersection is NOT [c,b] or necessarily (c,b) because what if d<b, then the intersection is in fact (c,d) and not (c,b) like you're assuming.

6. Sep 9, 2006

### HallsofIvy

Staff Emeritus
If you want to do well in Analysis, you are going to have to make an effort to at least write things correctly!. (a, b) is not the same as [a,b] and you are confusing the two.

No, you haven't- for two reasons. First, as others said, the intersection of the two open intervals, (a,b) and (c,d) cannot be the closed interval (c,b)- you are being careless with your notation. More importantly, what if a= 0, b= 10, c= 1, d= 2- that is, I= (0, 10) and J= (1, 2). Then the intersection is (1, 2), not (1, 10).

7. Sep 9, 2006

### benorin

There are only four possibilities that produce a non-empty intersection, namely:

a < c < b < d
a < c < d < b
c < a < b < d
c < a < d < b

now convince yourself that $$I\cap J = (a,b)\cap (c,d) = \left( \max \{ a,c\} , \min \{b,d\} \right)$$

8. Sep 9, 2006

### JasonRox

I never said that!

9. Sep 9, 2006

### JasonRox

That's what I was thinking.

10. Sep 9, 2006

### HallsofIvy

Staff Emeritus
Never said what?

In your first post you definitely did say

11. Sep 9, 2006

### matt grime

No he didn't. Zoix wrote it. Just scroll upwards....

12. Sep 9, 2006

### ZioX

I wasn't being careless in notation. I do know the difference between closed and open intervals, and their specific notation. I did indeed use notation for both open and closed intervals in my OP. I made the mistake that I thought the intersection would be a closed interval (which is why I used closed interval notation)...since they overlap they would have both, but I realize the error in this. I hope you realize your error too.

13. Nov 22, 2007

Just falling on this I thought it was funny. Actually, I think there are 9 possibilities of nonempty intersection:

a < c < b < d
a < c < d < b
c < a < b < d
c < a < d < b
a = c < b < d
a = c < d < b
a < c < b = d
c < a < b = d
a = c < b = d

Still get $$I\cap J = (a,b)\cap (c,d) = \left( \max \{ a,c\} , \min \{b,d\} \right)$$