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.