Let $(X,d)$ be a metric space. For every non-empty subset $E$ of $X$, let $\rho_E: X \rightarrow\Bbb{R}$ be the function defined by $\rho_E(x) = \inf\{d(x,E)\,| \, y \in E\}$. This function is uniformly continuous for every $E$.

Now let $F, K$ be closed disjoint subsets of $X$. I am thinking about the function $f: X \rightarrow \Bbb{R}$ defined by $$f(x)=\frac{\rho_F(x)}{\rho_F(x)+\rho_K(x)}, \quad \forall x \in X$$ that we use to prove metric spaces are perfectly normal. $f$ is continuous on $X$ since the denominator never vanishes. If it were not continuous, we could not use such a function to prove metric spaces are $T_6$. What about uniform continuity?

I found out that $f$ is not uniformly continuous for some combination of $F$ and $K$. For example, let $X = \Bbb{R}$, $F = \Bbb{N}^+$ and $K = \{n+1/n\,|\, n \in \Bbb{N^+}\}$. We can spell out the negation of "$f$ is uniformly continuous": Let $\epsilon = 1$. Pick any $\delta >0$. There is $N \in \Bbb{N^+}$ such that $1/N<\delta$. Take $x = N$ and $y = N+1/N$. $|x-y| = 1/N < \delta$ but $|f(x)-f(y)| = |0-1| = 1 \geq \epsilon$.

However, I also know that if $K$ is compact and $F$ is closed, then $K$ and $F$ must be separated by some positive distance. Essentially I cannot reuse the above argument. So that my question is,

If $K$ is compact, $F$ is closed, and $K \cap F = \emptyset$, is $f$ uniformly continuous on $X$?

Edit: I missed the most important assumption that $K$ and $F$ are disjoint. Now I added it back. Sorry about that.

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    $\begingroup$ Try showing that it is uniformly continuous if and only if the sets are a positive distance apart. $\endgroup$ – Daniel Fischer May 24 '17 at 11:17
  • $\begingroup$ @DanielFischer thanks! At least guessed correctly the way to go. (But I didn't continue due to lack of mentality. My bad. $\endgroup$ – Li Chun Min May 24 '17 at 11:28

Here is a general statement:

Claim. Suppose that $g, h : X \to [0, \infty)$ are uniformly continuous and $$ \delta := \inf_{x\in X} (g(x) + h(x)) > 0, $$ then $f(x) = \frac{g(x)}{g(x) + h(x)}$ is uniformly continuous.

The proof is quite straightforward:

\begin{align*} |f(x) - f(y)| &= \frac{|g(x)h(y) - g(y)h(x)|}{(g(x)+h(x))(g(y)+h(y))} \\ &\leq \frac{|g(x) - g(y)|h(y)}{(g(x)+h(x))(g(y)+h(y))} + \frac{|h(x) - h(y)|g(y)}{(g(x)+h(x))(g(y)+h(y))} \\ &\leq \frac{1}{\delta}(|g(x) - g(y)| + |h(x) - h(y)|). \end{align*}

Now can you check that this claim is applicable to your function if $\operatorname{dist}(F, K) > 0$?

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    $\begingroup$ Thanks! I didn't know how to deal with the ugly fraction and your answer exactly addressed the problem that I had. $\endgroup$ – Li Chun Min May 24 '17 at 11:30

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