Let $(X, d)$ be a metric space and $E ⊂ X$ be a nonempty subset. Define a function
$f : X → [0, ∞)$ by: $$f(x):= \inf \{d(x,y):y \in E \}$$
Prove that $f$ is uniformly continuous on $ X$ .
Let me share a very short solution that is confirmed by my instructor, but which I don't understand how the Triangle Inequality was applied.
Solution:
Let $\delta=\frac {\epsilon}{2}$ for any $\epsilon>0$
Let arbitrary points $x_1, x_2$ satisfy $d(x_1,x_2)< \delta$.
Then, $f(x_1) <f(x_2)+2 \delta$ and $f(x_2)<f(x_1)+ 2\delta$.
Thus, $|f(x_1)-f(x_2)|<2\delta=\epsilon$.
Hence it is uniformly continuous on $X$.
The instructor said, "$f(x_1) <f(x_2)+2 \delta$ and $f(x_2)<f(x_1)+ 2\delta$" could be found using the Triangle Inequality, but I wonder how. Can I get some enlightenment here?
You have that $x_1,x_2$ satisfy $d(x_1,x_2) < \delta$. Also, $$f(x_1):= \inf \{d(x_1,y):y \in E\}$$ $$f(x_2):= \inf \{d(x_2,y):y \in E\}$$ By the triangle inequality, $$d(x_1,y) \le d(x_1,x_2) + d(x_2,y)$$ for every $y$. Also, $$\inf\{d(x_1,y): y\in E\} \le d(x_1,y)$$ so, $$\inf\{d(x_1,y): y\in E\} \le d(x_1,x_2) + d(x_2,y)$$ for every $y$. Now the LHS is independent of $y$, and the inequality holds for all $y$. Taking the infimum over $y\in E$ we get $$\inf\{d(x_1,y): y\in E\} \le d(x_1,x_2) + \inf\{d(x_2,y): y\in E\}$$ but this is just $$f(x_1) \le d(x_1,x_2) + f(x_2)$$ and $d(x_1,x_2) < \delta$ quickly gives $$f(x_1) \le f(x_2) + \delta$$ You can just swap $x_1$ and $x_2$, repeat the entire procedure above to get $$f(x_2) \le f(x_1) + \delta$$ I hope this provides the necessary enlightenment.