# Metric space property of balls

Is it true that for a complete metric space $$X$$, for every continuous function $$\epsilon:X\rightarrow (0,\infty)$$, there exists a continuous function $$\delta:X\rightarrow (0,\infty)$$ such that for any $$x'\in X$$ if $$x\in B_{\epsilon(x')}(x')$$ then $$B_{\delta(x)}(x)\subseteq B_{\epsilon(x')}(x')$$?

Ideas:As $$B_{\epsilon(x')}(x')$$ is open, for any $$x\in X$$, there exists $$\delta_x>0$$ such that $$B_{\delta_x}(x)\subseteq B_{\epsilon(x')}(x')$$. But $$\delta:B_{\epsilon(x')}(x')\rightarrow (0,\infty)$$ given by $$\delta(x)=\delta_x$$ is only defined locally and don't have to be continuous. If $$\delta$$ is continuous, then maybe partitions of unity will give what we want?

• Let $\varepsilon:(0, \infty)\to (0, \infty)$ be $\varepsilon(t) = t$. Then the open balls $B(t, \varepsilon(t))$ are sets $(0, a)$ for some $a > 0$. But if $x\in (0, \infty)$ and $s > 0$, then $x\in (0, x+s/2)$ but the ball around $x$ with radius $s$ isn't contained in $(0, x+s/2)$. Commented Aug 13 at 20:58
• What I'm saying that you can't find such $\delta$ at all. Continuous or not. Commented Aug 13 at 21:00
• Your idea is wrong/incomplete by the way. You need $B_{\delta_x}(x)$ to be contained in $\bigcap_{x'} B_{\varepsilon(x')}(x')$ where the intersection goes over all $x'$ such that $x\in B_{\varepsilon(x')}(x')$. So you need $x$ to belong to interior of it. Commented Aug 13 at 21:06
• If you want $X$ to be complete then take $X = [1, \infty)$ and $\varepsilon:X\to (0, \infty)$ as above. Then open balls $B(t, \varepsilon(t))$ are sets $[1, a)$ for $a \geq 2$. Take $x = 2$ for example, and any $s > 0$. Then $x\in [1, x+s/2)$ but the ball around $x$ with radius $s$ isn't contained in $[1, x+s/2)$. Commented Aug 13 at 21:09
• So even if $X$ is complete, then can be no such $\delta$ at all. Commented Aug 13 at 21:10

Here $$B(x, r) = \{y\in X : d(x, y) < r\}$$. Let $$X = \mathbb{R}$$ with Euclidean metric and $$\varepsilon:X\to (0, \infty)$$ be defined as $$\varepsilon(t) = \max(1, t)$$.
Suppose that there exists a function, not necessarily continuous, $$\delta:X\to (0, \infty)$$ such that $$B(t', \delta(t'))\subseteq B(t, \varepsilon(t))$$ for any $$t'\in B(t, \varepsilon(t))$$ and $$t\in X$$.
Let $$s = \delta(2)$$. Then $$2\in B(1+\frac{s}{4}, \varepsilon(1+\frac{s}{4})) = (0, 2+\frac{s}{2})$$ but $$B(2, s) = (2-s, 2+s)$$ is not contained in $$(0, 2+\frac{s}{2})$$ since $$2+\frac{s}{2}\in (2-s, 2+s)$$ but $$2+\frac{s}{2}\notin (0, 2+\frac{s}{2})$$.
So such function $$\delta$$ cannot exist, continuous or not.