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The proposition on GP Page 203 says:

Let $S$ be a rectangular solid and $S_1, S_2, \ldots$ a covering of its closure of $\bar{S}$ by other solids. Then $\sum$vol$(S_j) \geq$ vol($S$).

This does not quite make sense to be - why can we assert that Volum of the covering of $\bar{S} \geq S$? I think I must missed something. One thing I am not sure about is thatis defined to be the whole space?

Thank you very much for helping me clear my confusion.

1LiterTears
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    I'm not sure what you're asking. The proposition is followed by a page-long proof. "One thing I am not sure about is thatis defined to be the whole space?" I'm not sure what this sentence means :( – Tim kinsella Jun 10 '13 at 17:10
  • Hi @Timkinsella, thank you for you comment. I understand $\bar{S}$ as the compliment of $S$. Hence, I am confused with what is the whole set - in other words, what is $S + \bar{S}$. Thanks. – 1LiterTears Jun 10 '13 at 17:13
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    Ohhh! It's the topological closure! :) – Tim kinsella Jun 10 '13 at 17:14
  • @Timkinsella, oh no. I got it - I confused closure with compliment! Thank you so much for helping me clear this out!! – 1LiterTears Jun 10 '13 at 17:14

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Thanks to @Timkinsella. This is trivially true, since the closure of a set certainly is greater than or equal to the set.

1LiterTears
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