I recently came across the following question:
Let $M$ be a metric space, $\mathcal{U}$ an open cover of $M$. For each $x \in M$, define $r(x):= \sup \{0 < r \leq 1 | \exists U \in \mathcal{U} \text{ such that } \mathbb{B}(x,r) \subset U \}$.
Is $r(x):M \mapsto (0,1]$ necessarily continuous?
After a small amount of thought, I guessed it is, and here is my proof:
Fix $x \in M$. Let $\epsilon > 0$.
Let $R := r(x) = \sup \{0 < r \leq 1 | \exists U \in \mathcal{U} \text{ such that } \mathbb{B}(x,r) \subset U \}$.
Define $\delta = \frac{\epsilon}{4} > 0$. Take $y \in \mathbb{B}(x,\delta)$.
We know as $R$ is a least upper bound that $\exists U \in \mathcal{U} \text{ such that } \mathbb{B}(x,R - \frac{\epsilon}{2}) \subset U$.
So for $z \in \mathbb{B} (y,R - \frac{3\epsilon}{4})$ , $d(z,x) \leq d(z,y) + d(x,y) < R - \frac{3\epsilon}{4} + \frac{\epsilon}{4} = R - \frac{\epsilon}{2}$.
So $\mathbb{B} (y,R - \frac{3\epsilon}{4}) \subset \mathbb{B}(x,R - \frac{\epsilon}{2}) \subset U$.
Therefore, $r(y) > R - \epsilon$.
Now for the other bound, assume $r(y) \geq R + \epsilon$.
Then, in a similar vein to the above, $\exists U \in \mathcal{U} \text{ such that } \mathbb{B}(y,R + \frac{\epsilon}{2}) \subset U$.
So for $z \in \mathbb{B} (x,R + \frac{\epsilon}{4})$ , $d(z,y) \leq d(z,x) + d(x,y) < R + \frac{\epsilon}{4} + \frac{\epsilon}{4} = R + \frac{\epsilon}{2}$.
So $\mathbb{B}(x,R + \frac{\epsilon}{4}) \subset \mathbb{B}(y,R + \frac{\epsilon}{2}) \subset U$.
So either $R = 1$, and then $r(y) \geq 1 + \epsilon$ which is a contradiction, or $R$ was not actually a least upper bound, which is also contradictory.
Therefore $R - \epsilon < r(y) < R + \epsilon$.
So $r(x)$ is continuous.
Does this seem correct? Is there a smoother way to do it? Or have I completely missed something and someone can provide a counterexample?!
Thanks.
For $U$, $f_U(x)=0$ for $x\in U^c$ and $f_U(x)=\min\ \{1, \sup\ \{r|B(x,r)\subset U\} \}$. Then $f_U$ is continuous. Further, $F(x)=\sup\ \{ f_U(x)|U$ is in cover$\}$ is continous.