# Intersection of two open sets

1. Aug 23, 2007

### tburnum

[SOLVED] intersection of two open sets

does anyone know how to prove the intersection of two open sets is open?

2. Aug 23, 2007

### cristo

Staff Emeritus
Yes, do you? What is the definition of an open set?

3. Aug 23, 2007

### scarecrow

(....) ∩ (....) = (.......)?

lol

4. Aug 24, 2007

### dextercioby

There's nothing to prove about it. The axioms of a topological space (topology over a set) claim that the intersection of an arbitrary but finite # of open sets is open.

Last edited: Aug 24, 2007
5. Aug 24, 2007

### matt grime

One presumes the OP is talking about the usual notion of open and closed in R. There is something to prove then if one wishes to show this is actually a model of the axioms of a topological space, dexter.

But, this does show the place to start is with the OPs definition of open (or for that matter, in what space are we even talking? R, R^n?)

6. Aug 24, 2007

### tburnum

the open sets are in R, but i need to prove that the intersection of just two open sets is open. once i have that, proving the intersection of a finite number of open sets is easy. i'm trying to use an open ball in the proof.
i'm at a loss.

7. Aug 24, 2007

### matt grime

Let's look at the problem. Let's take (a,b) and (c,d) as the open intervals. What is the intersection of those two sets? There are only a few cases to consider - they're disjoint, one is a subset of the other, and they overlap.

If you want to be fancier, and allow for arbitrary open sets, then what does it mean for X to be open? It means for each x in X there is an e depending on x such that (x-e,x+e) is wholly contained in X. Now suppose that x is in the intersection of X and Y. That means what? That there is an e such that (x-e,x+e) lies wholly in X, and a d so that (x-d,x+d) lies wholly in Y. Can you now see how to find an f so that (x-f,x+f) lies wholly in X and wholly in Y (i.e. in the intersection thus proving that the intersection is open)?

8. Aug 24, 2007

### tburnum

i have been able to prove that there is an e such that (x-e,x+e) is wholly in X and there is an f such that (y-f,y+f) is wholly in Y, i'm having a problem proving that there exists a g such that (z-g,z+g) exists only in the intersection of X and Y. i can't figure out what g should equal to make sure that the it stays in the intersection.

9. Aug 24, 2007

### matt grime

You can't? Well, what if I said x=0 which I may as well do. Then what if e=1 and g=2? Can you really not think of an open interval contained in both (-1,1) and (-2,2)?

And why do you have x,y, and z? You only have to suppose you are looking at one point x lying in both X and Y.

10. Aug 24, 2007

### tburnum

i am having a major brain fart. ok, so if i'm only looking at one point x that is an element of the intersection of X and Y, and e>0.
then the i have to consider the cases:
x is an element in X or x is an element in Y and they do not over lap, then the intersection is obviously open because the intersection is the empy set
so the case i need is when X n Y does not equal the empty set.
got it and i know how to prove an open ball is a subset of X n Y, i just can't figure out what i need my e=?

11. Aug 24, 2007

### HallsofIvy

If x is in open set A, there exist $\delta_1> 0$ such that $(x-\delta_1,x+\delta_1)$ is a subset of A. If x is in open set B, thre exist exist $\delta_2> 0$ such that $(x-\delta_2,x+\delta_2)$ is a subset of B. Now, how can you find a $\delta> 0$ so that both of those are true?

12. Aug 24, 2007

### tburnum

i'm trying to use the definition of intersection in my proof, but i can't seem to get it right. i can prove the x is an element in X or Y, but it's the and that's tripping me up.

13. Aug 24, 2007

### matt grime

I'm goint to repeat my question. Are you saying that you cannot see an open interval that is contained in (-1,1) and (-2,2).

Can you see one that is a subset of (-10,10) and (-0.1, 0.1)?

14. Aug 24, 2007

### tburnum

see yes, prove no

15. Aug 24, 2007

### matt grime

What open set are you thinking of? Write an open subset of (-1,1) and (-2,2) out for me.

16. Aug 24, 2007

### tburnum

i have to prove it for any X and Y. i has to be abstract, all theory. which my prof told me i have to follow the logic and i'm stuck.

17. Aug 24, 2007

### EnumaElish

You'll never get to abstract logic if you cannot solve a simple, concrete example.

If you want to be unstuck I suggest you try to work through a simple example like the one above.

Last edited: Aug 24, 2007
18. Aug 24, 2007

### tburnum

it's the number line with shaded between -1 and 1 with circles at the end to indicate that the end points are not included in the set

i'm thinking more of a Venn Diagram for a picture.

19. Aug 24, 2007

### EnumaElish

Stay with the example. What is your e, f and g for x = 0?

Last edited: Aug 24, 2007
20. Aug 24, 2007

### tburnum

i have e>0 and f>0, and i tried g=min{e,f} but that doesn't work because it doesn't guarantee that g will be small enough to only exist in the intersection of X and Y

21. Aug 24, 2007

### EnumaElish

Stay with the example. Can you write out an e, an f, and a g for x=0 in (-1,1) within (-2,2)?

22. Aug 24, 2007

### EnumaElish

Can you give an example on R where min doesn't work?

23. Aug 24, 2007

### tburnum

e exists in (-1,1) and =.9999999999.....
f exists in (-2,2) and =1.999999999......
in this case, if g=min{e,f} then it works
i get that, but what i don't get is when one is not a subset of the other and they just overlap a little (e.g. (-3,1) and (0,4)

24. Aug 24, 2007

### EnumaElish

Can you give an example with (-3,1) and (0,4) where min does not work?

Note: You cannot use x=0 because it is not in the intersection, so use any other x.

Last edited: Aug 24, 2007
25. Aug 24, 2007

### matt grime

GAH! Look, what is your definition of open?

I don't know why I'm doing this since it should not be necessary.

Or just forget that. Take two open intervals (a,b) and (c,d) that intersect, and suppose that one is not inside the other. Well, that means that a<c<b<d. But what is the intersection of the two sets? (c,b) and that's an open interval.

Now that ought not to help, because it is immaterial since the definition of open is NOT that it is an interval. It is that for any element x in X, there is an interval (x-e,x+e) about x contained in X. So you are completely missing the point if you're worrying about intervals not overlapping: either (x-e,x+e) is contained in (x-d,x+d) or vice versa.

Last edited: Aug 24, 2007