First question here on Math.stackexchange. I am working on a proof and looking for some help/guidance.
The question is as follows:
Let $A \subset\mathbb R^n$ be any nonempty set. Show that the set $B = \{\hat{x} \in\mathbb R^n \mid \operatorname{dist}(\hat{x},A) > 0\}$ is open.
My proof thus far:
1: I start by assuming A is a closed set.
If so is the case, then the closest point to any $\hat{x}$ will be a limit point on A. Since the distance between A and $\hat{x}$ is greater than $0$, can I instantly deduce that in this case B is an open set? (Since B would contain a Ball with a radius $> 0$ for any $\hat{x}$ and by definition be an open set.)
2: Now, assuming A is an open set:
In order to show that B is also an open set, B must contain a ball $C$, centered at $\hat{x}$ with radius $r > 0$ and points $\hat{d}$ around $\hat{x}$ , such that: $C_{r}(\hat{x}) = \{\hat{d} ∈ \mathbb R^n: 0 < |\hat{d}-\hat{x}|<r, r>0\}. $
We also know that $\hat{x}$ is of distance $dist(\hat{x},A) >0$ from A and we know that $\hat{x}$ is of some distance from $\hat{d}$, where $\hat{d}$ can be on the straight line between A and $\hat{x}$.
Also, since A is open, its' limit points is not in the open set A and can be said to be of distance $\epsilon, \epsilon > 0$.
If I now say that the distance between $\hat{x}$ and $A$, i.e. $dist(\hat{x},A) = |\hat{d}-\hat{x}| + \epsilon$. Then by setting $|\hat{d}-\hat{x}| = \frac{\epsilon}{2}$, we have the distance to be:
$dist(\hat{x},A) = \frac{\epsilon}{2} + \epsilon$.
Since the $dist(\hat{x},A)$ is greater than the distance $|\hat{d}-\hat{x}|$, this should prove that the ball C is in the set B?
Is this proof correct?