>>13759075Imagine you have x = 0 on the real line, and epsilon is 2, while y = 5.
The distance from y to x is 5, but the distance from y to the edge of the epsilon ball around x is 3 (draw it if you need). So that other ball is just barely outside of the ball around x (i.e. "not meeting", or "disjoint").
What this shows is that around every y that is more than epsilon away from x, there is an open ball containing y (and disjoint from the first one), so the set of all these points that are more than epsilon away is open (since it's a union of open balls around the y's). So its complement is closed, which is the set of points that are <= epsilon away from x.
So this is a closed set containing the epsilon ball, so it contains the closure (since the closure of a set S is the smallest closed set containing S).
>>13759109This is really a set theory question. You have to remember the property that preimage of the complement is the complement of the preimage (the two operations commute). Think about that for a bit.
