Closed sets

May 2009
86
1
Hey, I was just trying out some questions from Rudin, when I had this doubt as to whether my answer is right, can you please tell me ifI am right?

Let E' be the set of all limit points of a set E. Prove that E' is closed.

Let x be a limit point of E'. So every nbd of x contains infinitely many points of E'. But in every nbd of every point of E' are infinitely many points of E, since every point of E' is a limit point of E.
Thereby every nbd of x contains infinitely many points of E. So x is a limit point of E (ie) x is in E'.
Hence E' is closed. Am i right?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Hey, I was just trying out some questions from Rudin, when I had this doubt as to whether my answer is right, can you please tell me ifI am right?

Let E' be the set of all limit points of a set E. Prove that E' is closed.

Let x be a limit point of E'. So every nbd of x contains infinitely many points of E'. But in every nbd of every point of E' are infinitely many points of E, since every point of E' is a limit point of E.
Thereby every nbd of x contains infinitely many points of E. So x is a limit point of E (ie) x is in E'.
Hence E' is closed. Am i right?
This is true in any \(\displaystyle T_1\) space, but since it's Rudin it must be that you are speaking of a metric space.

What you said is pretty much it, but I think you need to say it slightly better. Every neighborhood \(\displaystyle U\) contains a point \(\displaystyle y\) of \(\displaystyle E'\) different from itself but since \(\displaystyle U\) is also a neighborhood of \(\displaystyle y\) it follows (since this is a metric space) that \(\displaystyle U\) contains infinitely many points of \(\displaystyle E\), in particular it must contain a point of \(\displaystyle E\) distinct from \(\displaystyle x\) and thus \(\displaystyle x\) is a limit point of \(\displaystyle E\) so \(\displaystyle x\in E'\)
 
Sep 2009
64
23
Edinburgh, UK
I think it is best just to show that the complement of \(\displaystyle E^{\prime}\) is open. So let the surrounding topological space (or metric space) be denoted by \(\displaystyle X\) and assume that \(\displaystyle E^{\prime}\subset X\) is a proper subset. Then let \(\displaystyle x\in X\backslash E^{\prime}\), so that \(\displaystyle x\) is not a limit point of \(\displaystyle E\). Then by definition there must be a neighbourhood \(\displaystyle U\) of \(\displaystyle x\) such that \(\displaystyle U\cap E^{\prime}= \emptyset\), i.e. \(\displaystyle x\in U \subseteq X\backslash E^{\prime}\). Since this holds for any such \(\displaystyle x\in X\backslash E^{\prime}\), this means that \(\displaystyle X\backslash E^{\prime}\) is open, so that \(\displaystyle E^{\prime}\) is closed.
 
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Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Then by definition there must be a neighbourhood \(\displaystyle U\) of \(\displaystyle x\) such that \(\displaystyle U\cap E^{\prime}= \emptyset\), i.e. \(\displaystyle x\in U \subseteq X\backslash E^{\prime}\).
That isn't true at all. It is true that there is a neighborhood such that \(\displaystyle U\cap E\subseteq\{x\}\)
 
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