- #1

Gear300

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If S = open interval (a,b), I don't really see how this could be the case (there will always be points in S that are not in the union of the disjoint sets).

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- Thread starter Gear300
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- #1

Gear300

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If S = open interval (a,b), I don't really see how this could be the case (there will always be points in S that are not in the union of the disjoint sets).

- #2

TylerH

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- #3

Gear300

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I don't know. Perhaps its a trivial instance in this case: S U {}, a union with a null set, in which the null set is both open and closed. Though, I'm not sure if this is what they have in mind.

- #4

HallsofIvy

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The "meat" of the theorem is that open sets that are NOT intervals can be written as unions of open

- #5

Gear300

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1. The example you are asking about, (a, b),isan open interval and so is the union of1open set- itself.

The "meat" of the theorem is that open sets that are NOT intervals can be written as unions of openintervals.

That makes things more clear. Thanks.

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