2
$\begingroup$

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?

$\endgroup$
10
  • 1
    $\begingroup$ 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 $(0, x+s/2)$. $\endgroup$
    – Jakobian
    Commented Aug 13 at 20:58
  • 1
    $\begingroup$ What I'm saying that you can't find such $\delta$ at all. Continuous or not. $\endgroup$
    – Jakobian
    Commented Aug 13 at 21:00
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jakobian
    Commented Aug 13 at 21:06
  • 1
    $\begingroup$ 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 $[1, x+s/2)$. $\endgroup$
    – Jakobian
    Commented Aug 13 at 21:09
  • 1
    $\begingroup$ So even if $X$ is complete, then can be no such $\delta$ at all. $\endgroup$
    – Jakobian
    Commented Aug 13 at 21:10

1 Answer 1

Reset to default
4
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .