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Covering Lemma: There exist $c\in \mathbb{N}$ ($c=2$ works) s.t. for any finite collection of open intervals $\{I_n\}$ there is a sub collection $\{I_{n_j}\}$, s.t. $\{I_{n_j}\}$ covers same set that $\{I_n\}$ cover and $\{I_{n_j}\}$'s can overlap at most $c$ times at any point. But it doesn't hold for squares in $\mathbb{R}\times\mathbb{R}$.

That is a homework question but I couldn't exactly understand what I should prove. What I think is since the collection is already finite they can only overlap finitely many times and that is why it seems obvious.I think I did not understand what the question exactly asks. Please some clarification!

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    The crucial point is that $c$ is supposed to be independent of the number of intervals.2012-11-03

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Since it says $c=2$ works you can prove the following:

For any finite collection of open intervals $\{I_n\}$ there is a subcollection $\{I_{n_j}\}$, s.t. $\{I_{n_j}\}$ covers same set that $\{I_n\}$ covers and $\{I_{n_j}\}$'s can overlap at most 2 times at any point.

EDIT: So, you should show that there is a suitable subcollection such that for any point $x\in\bigcup\{I_{n_j}\}$, at most 2 intervals in the subcollection contain $x$.

For squares in $\mathbb R\times\mathbb R$, if you take any $c\in\mathbb N$, there exists a finite collection of $c+1$ squares and a point which is common to every square (so at one point they overlap $c+1$ times) and you cannot remove a single square without changing the union of the squares. Thus, the lemma doesn't hold here.

EDIT: An example: The collection $\{(0,4)\times(0,4), (1,5)\times(1,5), (2,6)\times(2,6)\}$ of squares shows that $c=2$ is not enough for squares. Now these squares overlap three times at point $(3,3)$ but you can't remove any square without changing the covered set. You can generalize this for every $c$ and hence show that the lemma doesn't hold for squares.

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    Thanks a lot for the help.2012-11-03