Prove if S is an open set, then ~S is closed; if S is a closed set, then ~S is open.?

Definitions:

S is Open- for every point in S, there exist a segment r such that p in r and r subset S

S is Closed- if p is a cluster point of S, then p in S

Cluster point (p)- for each segment r containg p, there is a element q of S such q is in r and q not equal p

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  • 10 years ago
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    Suppose S is open and let p be a cluster point of ~S. We can establish that p is contained in ~S by contradiction. Suppose p is not contained in ~S. Then p is contained in S. Since S is open, there exists a segment r containing p such that r is a subset of S. Now, r is a segment containing p, and as p is a cluster point of ~S, the intersection r ∩ ~S-{p} is nonempty. So r intersects ~S. But this contradicts r being a subset of S. Hence p is contained in ~S. As p was arbitrary, we conclude that every cluster point of ~S is contained in ~S which is therefore closed.

    Conversely, suppose that S is closed and let p be any element of ~S. As p is an element of ~S, p is not contained in S and so p is not a cluster point of S. It follows that there exists a segment r containing p such that r ∩ S is empty. Thus r is a subset of ~S. Since this holds for every element p of ~S, we conclude that ~S is open. ■

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