Showing two epsilon balls intersection is empty

In summary, for any two distinct points x and y in a normed linear space X, there exists a positive number r such that the epsilon balls B(x,r) and B(y,r) have no intersection. This can be proven constructively by choosing a radius smaller than half the distance between x and y, or by contradiction by assuming the intersection is non-empty and showing that this leads to a contradiction.
  • #1
Visceral
59
0

Homework Statement


Suppose [itex]x,y \in X[/itex] which is a normed linear space and [itex]x\neq y[/itex]
. Prove that [itex]\exists r>0[/itex] such that [itex]B(x,r) \cap B(y,r)=∅[/itex]


Homework Equations


Epsilon Ball

[itex]B(x,r)={z \in X:||x-z||<r}[/itex]

The Attempt at a Solution



So my attempt here is via contradiction and its not workout out. I assume the intersection of the balls around x and y are not empty, and try to use the fact that x is not equal to y and use some typical triangle inequality tricks to attempt to get a contradiction. I always end up getting what I assumed to be true.

For instance, if the intersection is not empty then there exists an element in both balls. This would imply that the 2*r>ε=||x-y||. However I can't contradict this. What am I doing wrong? Please give any suggestions if you can. I missed this problem on a test and I still can't figure it out.
 
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  • #2
So you've assumed, toward a contradiction, that for all [itex]r>0[/itex], [itex]B(x,r)\cap B(y,r)\neq\emptyset[/itex] and shown that this implies that for all [itex]r>0[/itex], [itex]2r>\|x-y\|[/itex].

What happens as [itex]r\rightarrow 0[/itex], what does this say about [itex]x[/itex] and [itex] y[/itex], and how does this contradict the assumptions of the problem?

For the record, you don't need to do a proof by contradiction here. There is a fairly simple constructive proof (i.e. you can actually find a specific r that works) using the triangle inequality. Try to figure out how that works.
 
  • #3
Visceral said:

Homework Statement


Suppose [itex]x,y \in X[/itex] which is a normed linear space and [itex]x\neq y[/itex]
. Prove that [itex]\exists r>0[/itex] such that [itex]B(x,r) \cap B(y,r)=∅[/itex]


Homework Equations


Epsilon Ball

[itex]B(x,r)={z \in X:||x-z||<r}[/itex]

The Attempt at a Solution



So my attempt here is via contradiction and its not workout out. I assume the intersection of the balls around x and y are not empty, and try to use the fact that x is not equal to y and use some typical triangle inequality tricks to attempt to get a contradiction. I always end up getting what I assumed to be true.

For instance, if the intersection is not empty then there exists an element in both balls. This would imply that the 2*r>ε=||x-y||. However I can't contradict this. What am I doing wrong? Please give any suggestions if you can. I missed this problem on a test and I still can't figure it out.

If x ≠ y, then ||x - y|| is a positive number, say d.

If you take r smaller than d/2, then no point in B(x, r) can also be in B(y, r).
 
  • #4
gopher_p- I see what you mean, and thank you for the help

Mark44-This is exactly what I just realized! Thanks

I think I got it. I let ||x-y||= ε and let the balls have radius of ε/3 and there is no intersection. That way was really incredibly easy... Can anyone give me a hint on how you can do this by contradiction? I got 7/10 points for this proof on a test and got stuck at the end. Apparently by the professors grading I was "on the right track" but missed some final trick.

Thanks again
 

1. What is the definition of an epsilon ball?

An epsilon ball is a set of all points within a specified distance (epsilon) from a given point. It is commonly used in mathematics and physics to define neighborhoods and open sets.

2. Why is it important to show that the intersection of two epsilon balls is empty?

Showing that the intersection of two epsilon balls is empty is important in proving the separation of two sets. It also helps to establish the non-existence of a common point between two sets, which can be useful in various mathematical proofs and applications.

3. How do you mathematically show that the intersection of two epsilon balls is empty?

To show that the intersection of two epsilon balls is empty, we can use the distance formula to calculate the distance between the centers of the two balls. If this distance is greater than the sum of the radii of the two balls, then the two balls do not intersect and the intersection is empty.

4. Can you give an example of how to prove that the intersection of two epsilon balls is empty?

Yes, for example, let's consider two epsilon balls with radii of 2 and 3, and centers at (0,0) and (5,0). The distance between the two centers is 5, which is greater than the sum of the radii (2+3=5). Therefore, the two balls do not intersect and the intersection is empty.

5. How does the concept of epsilon balls relate to the concept of open sets?

Epsilon balls are commonly used to define open sets. An open set is a set that contains all its interior points, meaning that for every point in the set, there exists an epsilon ball around it that is completely contained within the set. This means that every point in an open set has a neighborhood that is also contained in the set. Epsilon balls help to define these neighborhoods and establish the openness of a set.

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