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Say I have some set $E\subset \mathbb{R}$, and an open cover $\cup A_i \supset E$. Is there a general algorithm that makes this cover disjoint?

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In general, an open cover of $E$ does not have a disjoint subcover. What is meant by making "this" cover disjoint? –  André Nicolas Apr 15 '11 at 18:16
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I imagine you intended $E\subset R$, not $E \in R$. –  André Nicolas Apr 15 '11 at 18:26
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In addition to what user6312 wrote, you may want to look at the Vitali covering lemma. –  Willie Wong Apr 15 '11 at 18:28

1 Answer 1

You cannot make the cover pairwise disjoint in general (that is, with $A_i\cap A_j=\emptyset$ if $i\neq j$).

To see this, note that if you can express $E$ as a union of pairwise disjoint open sets, each of which has nonempty intersection with $E$, then $E$ is disconnected: just consider a single $A_i$, and the union $\cup_{j\neq i}A_j$ for a disconnection of $E$; that is, $E$ will have at least as many connected components as the cardinality of the index set.

So, for example, you cannot take an open cover for $[0,1]$ and turn it into a disjoint open cover, unless all but one of the sets don't actually intersect $[0,1]$, because $E$ is connected (has one connected component), so the index set of a pairwise disjoint open cover in which every element of the cover intersects $[0,1]$ must have cardinality at most $1$.

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