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I know that every subspace of $R^k$ is Lindelöf, i.e.: If G is a subspace of $R^k$, then any open covering of G has a countable sub-covering.

I was thinking whether it is true that, given G a subspace of $R^k$ any closed covering of G (covered by closed boxes, (coordinate of each dimension has the form [a,b], $(-\infty, a]$, or $[a,+\infty$)) has a countable sub-covering? in the case k=1, it seems true to me.

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    No: consider an uncountable set $G$ and the covering $\{\{x\}:x\in G\}$.2012-05-16
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    ah, Sorry, I should include that The covering has no degenerate set. For example in $R^k$, the set in the covering has dimension k?2012-05-16
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    How are you defining dimension?2012-05-16
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    What's about boxes instead. (closed boxes, or has a dimension with coordinate $[a,+\infty)or (-\infty, b])$. Do you think it's good now?2012-05-16
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    Yes, I’m pretty sure that it’s true if you cover with boxes with non-empty interiors, though I’d want to work through a proof to be sure. On the line it’s definitely true; in $\Bbb R^k$ with $k>1$ the proof will take a bit of work.2012-05-16

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In the plane, cover the closed unit disk by all closed rectangles with their corners on the unit circle.