1
$\begingroup$

Let $X$ be a metric space with $D \subseteq X$ a dense subset.

If there is a covering for $D$, under which conditions on the covering is it possible to guarantee that the covering also covers $X$?

For a simple example, if you have that the diameter of the elements of the covering is globally bounded below by a positive number, it is immediate that such a covering will cover the whole space.
And if the diameter is not necessarily bounded below, but for every sequence of elements of the covering with diameter converging to zero, every limit point in $X$ of the sequence is contained in a particular element of the covering, then the covering also covers $X$.

$\endgroup$
  • 1
    $\begingroup$ This was crossposted to MO. In the future, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see hear from you that you'd already gotten the solution elsewhere. $\endgroup$ – Zev Chonoles Jun 20 '13 at 6:06
1
$\begingroup$

A covering $\mathcal C$ of the set $D$ covers $X$, provided $\mathcal C$ has a Lebesgue number in $X$, that is there exists a number $\varepsilon>0$ such that for each point $x\in D$ there exists an element $C\in\mathcal C$ such that $C$ covers the ball $B_X(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.