Just take .
That single value works for both.
This is from a real analysis textbook so I'm not 100% sure if it should go in here or in analysis.
Show that if a,b belong in R, and a=/=b, then there exist epsilon-neighborhoods U of a and V of b such that the intersection of U and V equals to an empty set.
This is what I have so far:
Assume that there does not exist epsilon-neighborhoods U of a and V of b such that the intersection of U and V equals an empty set. Let, 0< epsilon < absolute value of (1/2(b-a)). Therefore, since the intersection of U and V does not equal an empty set, there must exist a y with which y-a < epsilon and y-b < epsilon.
Then I calculated the bounds of the union of all epsilon neighborhoods within the range of 0< epsilon < absolute value of (1/2(b-a)) giving U the bounds 1/2(3a - b) and 1/2(a+b); V the bounds 1/2(3b-a) and 1/2(a+b).
Am I on the right track? I know I have to somehow prove that if y-a < epsilon and y-b < epsilon then it cannot be within both of those bounds, but I don't how to do that...
So taking this new approach where, . I find that the two sets are:
To prove they are disjoint, I would have to find that no element of U can be found in V.
So I let .
Let us assume that , also.
Therefore, I have to prove that to find the 'contradiction'.
Or am I thinking about it too complicated again?