Set of limit points is always closed

**dropout_expert** · March 13th 2009, 01:12 PM

Problem: Let $E'$ be the set of all limit points of $E \subset X$ . Prove that $E'$ is closed.

I think I can do it if the set $E'$ has the least-upper-bound property.

Attempt: Let $p$ be a limit point of $E'$ . (If the set of $p$ is finite or empty then $E'$ is closed.) Then for every $r > 0$ there exists a point $q \ne p$ such that $q \in E'$ . Let $A$ be the set of points of $q \in E'$ such that $q<p$ . This is bounded above by $p$ , which is the lub of $E'$ . Hence $\text{sup}(A) = p$ and $p \in E'$ by the least-upper-bound property. Since $p$ was arbitrary, this proves that $E'$ is closed.

Firstly, does this sound okay? If so, how could I prove this without using the least-upper-bound property?

I don't have anybody to look at my work, so I may be making huge fallacies or assumptions, and I'd greatly appreciate any input on this. Thanks.

**Plato** · March 13th 2009, 01:40 PM

Before even reading what you have done, we must know basic facts.
What sort of space is X?
Is X a general topological space?
Is X a metric space?
Is X simply the set of real numbers? This is the only way you have LUB property.
Please give a complete answer.

**dropout_expert** · March 13th 2009, 01:42 PM

Sorry, X is a metric space.

**GaloisTheory1** · March 13th 2009, 02:11 PM

Consider a point in $(E')^c$ . Since $x$ is not in $E'$ , there must be an open set $U$ containing $x$ whose intersection with $E$ is either empty or $x$ itself. Either way, the intersection of $U$ with $E$ is empty. Hence, every $x \in (E')^c$ has a neighborhood that is a subset of $(E')^c$ . This means that $(E')^c$ is open and hence $E'$ is closed.

**Plato** · March 13th 2009, 02:13 PM

In a metric space $X$ if any point $p$ is a limit point of $E$ then every open set, $<br /> p \in O \;\& \;O \cap E\backslash \{ p\} \ne \emptyset$ .
In other words, any open set that contains a limit point must contain another point of the set.
So if $q \in \left( {E'} \right)^\prime \;\& \;q \in O\;\& \;\left( {\exists r \in \left( {E'\backslash \{ q\} } \right) \cap 0} \right)$ .
This being a metric space and $q \ne r$ this means for some other open set $r \in Q\; \subseteq O\backslash \{ q\}$ .
But that means that $r$ is a limit point of $E$ so $\left( {\exists s \in \left( {E\backslash \{ r\} } \right) \cap Q} \right)\; \Rightarrow \;s \in O \cap E\backslash \{ q\}$ .
That means that any open set containing $q$ contains a point of $E$ distinct from $q$ .
That proves that $q$ is a limit point of $E$ or $q \in {E'}$ so ${E'}$ is closed.

**HallsofIvy** · March 13th 2009, 04:21 PM

Originally Posted by dropout_expert

Sorry, X is a metric space.

Then you certainly can't use the least upper bound property because most metric spaces don't even have an order defined.

The complement of the set of limit points of a set is, of course, the set of non-limit points. If you can show that around every non-limit point, there is a neighborhood that includes no limit points, you will have shown that the complement to the set of limit points is open and so the set of limit points is closed.

I think you should be able to do that by looking at the negation of the definition of "limit point", which is the definition of "non-limit point".

**dropout_expert** · March 13th 2009, 04:42 PM

Thanks for the replies everybody.

Yes, I realise the lub property doesn't hold in every metric space, but I didn't know how to prove it in the general case. I was trying to prove a simpler case first (when X=R or any set with the lub property) and then see if I could extend that providing it was correct; of course, on reflection, it seems very futile to try to extend it if the proof relies on the lub property...

Apologies and thanks for you work.

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