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Originally Posted by pberardi  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