Thread: L(U) set of limit points of a set U in a metric space X is closed

Hey guys. I was looking at the negation proof of this on the forum, showing that there exists a neighborhood not containing limit points for every element in complement of L(U), which I can understand. I was just wondering though if this direct proof works as well.

L(U) is a set of limit points. Let x be a limit point of L(U). My claim is that x is also a limit pt of U.

Pf: nontrivially. Let z be a non trivial pt in that neighborhood. I want to show for every neighborhood around x I can find pts ,y, of U. Clearly by triangle inequality d(x,y) <= d(x,z) +d(z,y). Since z is a limit pt of U d(z,y) is arbitrary. Therefore we showed that d(x,z) and d(z,y) are arbitrary and that you can find neighborhoods of x s.t. it intersects U nontrivially. Thus x is also a limit point of U.

Critique? In case I forgot to consider something about the underlying metric space that I am defining neighborhoods with respect to. Then I feel the italicized portion is where it would trip my proof up.

How are you defining "closed set"? In Royden, for example, he defines a closed set as a set that equals its closure (or, a set that equals its limit points). The first proposition after defining "closed set" is that the closure of the closure of a set is equal to the closure of the set. (The operation of obtaining the closure of a set is idempotent.) It seems to me that this is the proposition you're trying to prove; is this correct?

Originally Posted by Ackbeet

In Royden, for example, he defines a closed set as a set that equals its closure (or, a set that equals its limit points).

One comment, the closure of a set is the union of the set with the set of limit points.
Note that not all points of the closure are limit point. In this problem .
In general spaces the set does not have to be closed.
However, this appears to be a metric space. So it is an Hausdorff space.
Thus is closed.
Note that is an open set and there is