cluster point?

Question

1. S is a nonempty subset of R that is bounded from above but has no greatest element (max). prove supS is a cluster point of S

2. Prove that a subset of a metric space is closed iff it contains all its cluster points.

Answer 1

1. According to the definition of supremum, for every eps >0 supS - eps is not an upper bound of S. Therefore, there exists s in S such that supS - eps < s < supS (since S has no greatest element, it's supremum is not in S, so that s < supS for every s in S). This shows every open interval containing supS (which is a neighborhood of supS) contains an element of S distinct from supS. Therefore, supS is a cluster point of S.

2. Suppose S is closed, so that it's complement S' is open. Then, every x of S' is an interior point of S' and, therefore, has a neighborhood V contained in S'. It follows V doesn´t intersect S, which shows no element of S' is a cluster point of S. It follows that, if y is a cluster point of S, then y is in S, proving the assertion.

On the other hand, if S contains all it's cluster points, then no x in S' is a cluster point of S. It follows x has a neighborhood V that contains no element of S distinct from x. Since x is in S', it follows V doesn't intersect S, that is, V is a subset o S'. Therefore, x is an interior point of S'. Since this holds for every x in S', we conclude S' is open, which implies S is closed.