Analysis on metric spaces 6) Prove that a set S in a metric space is closed if andonly if it cantains all of its accumulation points Analysis on metric spaces 6) Prove that a set S in a metric space is closed if andonly if it cantains all of its accumulation points Solution I assume you know that a set is closed if it is the complementof an open set. Suppose S is closed. Let x be any accumulation pointof S. If x is not in S, then it is in its complementwhich is open and hence contains an open ball Bcentered on x. Since x was the accumulation point so this ballalso contains an element y (yx) such that yis in S. This contradicts B being entirely within S\'scomplement. Conversely, suppose S contains all its accumulation points.Suppose the complement of S is T. We shall show that T is an openset. This will be sufficient to prove that S is a closed set. Let xbe an element of T. If all balls centered on x wereintersecting S then x would have been an accumulation point of S,hence inside S, hence not inside T. So at least 1 ball liesentirely within T. So T is open..

1. Analysis on metric spaces
6) Prove that a set S in a metric space is closed if andonly if it cantains all of its accumulation
points
Analysis on metric spaces
6) Prove that a set S in a metric space is closed if andonly if it cantains all of its accumulation
points
Solution
I assume you know that a set is closed if it is the complementof an open set.
Suppose S is closed. Let x be any accumulation pointof S. If x is not in S, then it is in its
complementwhich is open and hence contains an open ball Bcentered on x. Since x was the
accumulation point so this ballalso contains an element y (yx) such that yis in S. This contradicts
B being entirely within S'scomplement.
Conversely, suppose S contains all its accumulation points.Suppose the complement of S is T.
We shall show that T is an openset. This will be sufficient to prove that S is a closed set. Let xbe
an element of T. If all balls centered on x wereintersecting S then x would have been an
accumulation point of S,hence inside S, hence not inside T. So at least 1 ball liesentirely within
T. So T is open.