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I am reading this.

It says

Intuitively, an open set is a solid region minus its boundary. If we include the boundary, we get a closed set, which formally is defined as the complement of an open set.

Now, question is if a closed set includes interior points also then how can it be complement?

I know basic set theory. Enlighten me! :)

Thanks!

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    The closed set you get by including the boundary is not the complement of the previously mentioned open set. It is the complement of a different open set, namely the "outside" of the solid region.2010-12-31
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    Helpful comment thanks!2010-12-31

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A Closed set is by definition a set whose complement is an open set. Note that this also includes the possibility that a set is both open and closed, for example in a space with two connected components, each component is both open and closed.

Now, in what you have highlighted the complement of the solid region (inclusive of boundary) i.e. the whole space without the region, is open. Which, means that the solid region (inclusive of boundary) is closed.

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    How strange... when I was typing this a few seconds ago there were no replies and now there are two, posted more than 5 mins ago, is this a fault of my browser or some lag problems with the site?2010-12-31
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    I'm surprised you didn't get an alert about my answer being posted. I deleted it anyway, partly because it seemed redundant.2010-12-31
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    I don't understand. There was some answer earlier which I think I understood but its deleted. I don't know any topology(connected components?).2010-12-31
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    Yeah I didn't get an alert, but that is most probably because the answers were already posted while I was typing mine but the browser was not showing them. Perhaps this has something to do with chrome.2010-12-31
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    @TheMachineCharmer: A space is called disconnected if we can find two disjoint open sets that together form the whole space. In which case the complement of one is the other, therefore, each is both open and closed.2010-12-31
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    @TheMachineCharmer: For example, take the space you are considering and remove the boundary of the solid region from the space.2010-12-31
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    It makes sense now! Thanks! :)2010-12-31