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- Thread starter seema283k
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First, are you talking about an open-and-closed subset of R? If you are, then the empty set is also open-and-closed in R, so you have to specify that A is nonempty.

My approach for this proof would be to consider this.

An open set contains none of its boundary points.

A closed set contains all its boundary points.

The only way for this to be possible is for A to have NO boundary points at all. Show how {} and R are the only two sets that have this property.

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