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.
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.
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?