Show that $\{y \in X \mid d(y,x) > r\}$ is open for every $r \in \mathbb{R}$ $x\in X$
Show that $\{y \in X \mid d(y,x) > r\}$ is open for every $r \in \mathbb{R}$
My first thought was that what the distance between y and x, it is a closed set. Apparently, it is not. I'm trying to figure out, how to prove this. I haven't gone far because I don't know the intuition. Can anyone help me get started? 
 A: Let $U$ be the set in question. Then $B(y,{1 \over 2} (d(x,y)-r)) \subset U$,
hence $U$ is open.
Another way to look at it is to notice that $f(y) = d(y,x)$ is continuous
and $U = f^{-1} ((r,\infty))$. Since $f$ is continuous, the inverse image of an open set is open.
A: Define $O = \{y : d(y,x) > r\}$. To see that $O$ is open we will show that every $p \in O$ is an interior point of $O$.
Let $p$ be such that $d(x,p) > r$, for some fixed $r$. 
Then our "wiggle room" for $p$ is $\delta :=  d(x,p) - r > 0$.
Then $B(p, \delta) \subset O$, so $p$ is an interior point of $O$: 
let $q \in B(p, \delta)$.
Suppose that $d(q,x) \le r$, then we would have by the triangle inequality:
$$d(p,x) \le d(p,q) + d(q, x) < \delta + r = d(x,p)$$
which is a blatant contradiction. The last equality follows from the definition of $\delta$ and the second $<$ from $q \in B(p, \delta)$.
So $d(x,q) \le r$ is false, hence $q \in O$. 
As $q \in B(p, \delta)$ was arbitrary, we are done: all points of $O$ are interior points.
Note that this proof works in any metric space.
