Two questions about neighborhoods

pberardi · #1 $Old$ May 24th, 2009, 08:16 PM

I think I have the idea but I am having trouble finishing off these. Can I just post the questions and my attempt at the solutions hoping that someone can explain how to finish or if I am right? Thanks.

#1 Question
Let x be in R(real #s) and let E(epsilon) > 0. Prove that if y is in (x-E,x+E), then (x-E,x+E) is a neighborhood of y.

#1 Solution attempt
Given that y is in (x-E,x+E) we want to show by definition of neighborhood that there exists a positive number E s.t. (y-(x-E),y+(x+E)) is a subset of the interval (x-E,x+E). Since y is in (x-E, x+E), x-E<|y|<x+E. And y<x+E, y<-(x+E). At this point I think to myself that this is almost obvious. If y is inside the interval then of course the interval is a neighborhood of y. But then I think that then why did they ask the question. So I am a bit stuck.

#2
Let x, y be in R(real #s) with x != y. Prove that there is a neighborhood U and a neighborhood V s.t. the intersection of U and V is the empty set. i.e. U and V are disjoint sets.

#2 Solution attempt
Here I just pick E to be 1
Therefore x-1<x<x+1 and y-1<y<y+1
or: (x-1,x+1) is U and (y-1,y+1) is V
so |x| < 1 and |y| < 1. I thought I had it because I thought the the only way this could be true is if x = y. But then what if x was 1/2 and y was -1/2, this could still hold. So I thought maybe I am not thinking right.

Any help on these two?

Gamma · #2 $Old$ May 24th, 2009, 11:27 PM

1) I am not sure what your definition of neighborhood is, but when I learned it, it was just an open set that contains the point. By assumption $y\in (x-\epsilon , x + \epsilon )$ and this is a basis element for $\mathbb{R}$ so it is open. It doesnt matter that it is not centered at y.

2)If $x\not = y$ then let $\delta= |x-y|$ . Then consider $U=B(x, \frac{\delta}{2})$ and $V=B(y, \frac{\delta}{2})$ . These are the neighborhoods you are looking for.

NOTE: $B(x,\delta)=\{ y\in \mathbb{R}| |x-y|< \delta \}$

xalk · #3 $Old$ May 25th, 2009, 05:07 AM

Quote:

Originally Posted by pberardi $View Post$

I think I have the idea but I am having trouble finishing off these. Can I just post the questions and my attempt at the solutions hoping that someone can explain how to finish or if I am right? Thanks.

#2
Let x, y be in R(real #s) with x != y. Prove that there is a neighborhood U and a neighborhood V s.t. the intersection of U and V is the empty set. i.e. U and V are disjoint sets.

#2 Solution attempt
Here I just pick E to be 1
Therefore x-1<x<x+1 and y-1<y<y+1
or: (x-1,x+1) is U and (y-1,y+1) is V
so |x| < 1 and |y| < 1. I thought I had it because I thought the the only way this could be true is if x = y. But then what if x was 1/2 and y was -1/2, this could still hold. So I thought maybe I am not thinking right.

Any help on these two?

In those cases a geometrical picture always help.So take two different points x,y on the real No line .The distance between them is |x-y|.

SO if you define two neighborhoods with centers x and y respectively and radius |x-y|/3 ,then those two neighborhoods have no common points.

An analytical proof goes as follows:

Suppose z belongs to the U(x,|x-y|/3) ,a neighborhood round x,THEN ,|z-x|<|x-y|/3................................................. ...................................1

Now we must show that z does not belong to the V(y,|x-y|/3) ,a neighborhood round y
Assume it does.then:

|z-y|<|x-y|/3................................................. .........................2

Add (1) and (2) and we get:

|x-y|<|x-y|/3 a contradiction