Why is this set open? (Preimage)

Question

Let $(M,d)$ be a complete metric space and $A$ be a closed subset of $M$. Then the set \begin{equation} B:= \{x\in M: d(x,A) < 1/2\} \quad \text{is open}. \end{equation} ($d(x,A)= \inf_{y\in A} d(x,y)$)

Why would that be the case? If $B=\{x\in M:0<d(x,A)<1/2\}$ then one could argue that B is the preimage of an open set and thus be open since $d$ is continous as a metric. But the set above is also containing $x_0\in M$ such that $d(x_0,A)=0$. So that argument would'nt work right or am I missing something?

Answer 1

Here's another way.

To show that $B$ is open, we can take any $x \in B$ and show that there exists a ball $D (x, \varepsilon)$ such that $D (x, \varepsilon) \subseteq B$.

So, take arbitrary $x \in B$. Then we have $d (x, A) < \frac{1}{2}$. Consider the ball $C = D \left( x, \frac{1}{2} \left( \frac{1}{2} - d (x, A) \right) \right)$. We show $C \subseteq B$. For take any $x' \in C$. Then $d (x', x) < \frac{1}{2} \left( \frac{1}{2} - d (x, A) \right)$. By triangle inequality applied over the infimum,

$$d (x', A) \le d(x', x) + d(x, A) < \frac{1}{2} \left( \frac{1}{2} - d (x, A) \right) + d (x, A) = \frac{1}{2} \left( \frac{1}{2} + d(x, A) \right)$$

It follows from the above that $d(x', A) < \frac{1}{2} \left( \frac{1}{2} + d (x, A) \right) < \frac{1}{2} \left( \frac{1}{2} + \frac{1}{2} \right) = \frac{1}{2}$, so that $x' \in B$. Since $x'$ was arbitrary, we have that $C \subseteq B$. And since $x$ was arbitrary, $B$ is open.

Note that we did not have to use the fact that $A$ is closed. The proof still applies if $A$ is not closed.

Answer 2

For $A\subset M$ ( needn't be closed), $d_A:X\to\Bbb{R}$ defined by $$d_A(x) =d(x, A) $$

is continuous map ( even uniformly continuous, even Lipschitz map)

$B_1=d_A^{-1}\{(-\infty, \frac{1}{2})\} $

$B_2=d_A^{-1}\{(0, \frac{1}{2})\}$

Side note: For $A\subset M$ closed, $B_2=(M\setminus A) \cap B_1$.