# Need help understanding this problem

1. Jan 28, 2005

### dlee48

Let A be the set of real numbers x such that 0<x<=1. For every x element of A, let E sub x be the set of real numbers y such that 0<y<x. Then

for x in A the intersection of all Esubx is empty.

I do not understand that the intersection is empty. I see that my index set is infinite and that as x approaches zero the magnitude of the sets Esubx become infinitely small so that the intersection is not abtainable but does that mean it is non-exisitant?

Rudin suggests I note that for every y>0, y not in Esubx if x<y. Hence y not in the insection of all Esubx. I have a problem understanding this suggestion. My index set is bounded above and and not bounded below. It seems to me that y could be greater than or equal to x but not just greater than x. Vastly confused?!??!

2. Jan 28, 2005

### mathman

This might help. If x<=0 or x>1, x cannot be in the set in question, since it is not in the starter set. Consider any number u in the interval remaining, 0<u<=1. Let x=u/2, then u is not in Ex. Moreover Ex is one of the sets in the intersection, since x>0. Therefore u is not in the intersection. Since this is true for every u in the interval, there are no real numbers in the intersection set.

3. Jan 28, 2005

### dlee48

Well, it seems to that my index set A consists of real numbers (0,1]. How can we consider numbers outside of that interval. To me Esubx is (0,x], x<=1. It looks to me to be a magnatude. I can't keep from visualizing the Esubxs as magnatudes, as if you took all the bars in a bar chart and super imposed them all onto one bar and the littlest one would be the intersection of all of them. But of cousre, the littlest one is to little to find.

4. Jan 28, 2005

### hypermorphism

Your visualization is a countable infinity (each discrete bar can be given a natural index). Although your sets have been indexed, they have not been indexed discretely (they have been indexed with real numbers) so that you can group them as you believe above. When it comes to uncountable infinities, visualization sometimes yields nothing coherent at all (Ie., see Cantor's ternary set).
Instead, use what you know to be concrete properties of the real number system. If y, a real number, is in every Ex, then y < x for all x in the interval (0, 1]. Since y is also in the interval of question, y is the greatest lower bound of the interval. But the GLB of the interval is 0, and the GLB of a real set is unique. Thus y must be 0. But 0 is not an element of any Ex, thus y cannot be in the intersection of said sets. But y was an arbitrary element of the intersection, thus the instersection is the null set.

5. Jan 28, 2005

### dlee48

Yes, like I said the littlest one is to little to find so the magic is that y can't be in (0,1] hence the intersection is empty. Still a little confused about RUdin comment . Thanks.

6. Jan 28, 2005

### hypermorphism

Rudin went with a complementary approach to mine. He notes that since you can always find an x strictly less than every y in that interval, you could not choose a y in the interval that was less than every x in the interval.

7. Jan 28, 2005

### dlee48

Thanks again!

8. Jan 28, 2005

### dlee48

By the way, do you know how people are able to get all the math symbols into thier submission so they can use the symbol for intersection instead of the word.

9. Jan 29, 2005

### matt grime

Use latex, There is a explanation in physics.

From what you've written it doesn't seem that you understand what the example states, or what interesection means.

If X_s is a collection of sets labelled by s in S, some index set, then $$\cap X_s$$ is the set of elements that lie in all the X_s. So in the case in question, $$\cap X_s$$ s in (0,1] is clearly some subset of (0,1]. Suppose that the intersection is non-empty, and that t is an element in this intersection. Consider the set X_{t/2}, then t isn't in X_{t/2} so there is no such t in the intersection since it isn't in ALL the X_s (ie we've shown one (of the infinitely many) sets that do not contain it). t was arbitrary, thus it follows that the intersection must be empty.

10. Jan 29, 2005

### HallsofIvy

Staff Emeritus
The simplest way to see it is this: Suppose Y were in the intersection. Then it must be a positive number and in every Ex. But since Y is a positive number, there exist a positive number x such that x< Y (Y/2 for example). Y is NOT in Ex for THAT x. Therefore there can be no Y in the intersection: it is empty.

Notice that focusing on a specific Y is the key here.

11. Jan 29, 2005

### dlee48

Thanks, I understand perfectly after Hypermophisms comments.

12. Jan 29, 2005

### mathwonk

no positive number is closer to zero than all positive numbers.