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Let $X$ be a metric space with $D⊆X$ 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 $X$.

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    $\begingroup$ Crossposted from math.SE: math.stackexchange.com/q/425179/264 $\endgroup$
    – Zev Chonoles
    Commented Jun 20, 2013 at 6:06
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    $\begingroup$ If the diameters bounded below, the cvering of D might not cover X. For example, let X be the real line. D be the set of rational numbers and it is covered by two sets $(-\infty,\sqrt{2})$ and $(\sqrt{2},+\infty)$. $\endgroup$
    – Anton Petrunin
    Commented Jun 20, 2013 at 6:30
  • $\begingroup$ Diameters really have very little to do with it. Given any covering of $D$ that misses a point $p$ of $X$, and any set $S$ of diameter $d$ disjoint from $p$, you can take the unions of the members of the covering with $S$ and get a new covering whose elements all have diameter $\ge d$, but it still misses $p$. $\endgroup$
    – Robert Israel
    Commented Jun 20, 2013 at 7:48

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