# Thread: Proof: Accumulation points and closed sets.

1. ## Proof: Accumulation points and closed sets.

I need help with this question cause I don't know where to start off:

Prove that if a set contains each of its accumulation points, then it must be a closed set.

2. Originally Posted by Porter1
I need help with this question cause I don't know where to start off: Prove that if a set contains each of its accumulation points, then it must be a closed set.
What sort of space are you working in?
Is it a simple metric space, a Hausdorff space, or just a general topological space?
It does make a difference is the answer.

3. Oh sorry its just a general topological space

4. Originally Posted by Porter1
Oh sorry its just a general topological space
In that case it may not be true.
So what topology do you have on this space?

5. Originally Posted by Porter1
I need help with this question cause I don't know where to start off:

Prove that if a set contains each of its accumulation points, then it must be a closed set.
Originally Posted by Plato
In that case it may not be true.
So what topology do you have on this space?
Maybe I am overlooking something here, but I thought that in a topological space $X$ A set $E$ is closed iff $E=\bar{E}$ and $\bar{E}=E\cup D\left(E\right)$, where $D\left(E\right)$ is the derived set (set of all limit points of $E$, and so $D\left(E\right)\subseteq E\implies E=E\cup D\left(E\right)=\bar{E}$ and so $E$ is closed?

I am relatively new to general topology.

6. Originally Posted by Drexel28
Maybe I am overlooking something here, but I thought that in a topological space $X$ A set $E$ is closed iff $E=\bar{E}$ and $\bar{E}=E\cup D\left(E\right)$, where $D\left(E\right)$ is the derived set (set of all limit points of $E$, and so $D\left(E\right)\subseteq E\implies E=E\cup D\left(E\right)=\bar{E}$ and so $E$ is closed?
All of that is very true. But that is not the question.
The question is: Prove that the derived set of a set is closed. And that is not true.
Think of the space of real numbers, $\mathbb{R}$ with a topology generated by the collection $\left\{ {( - \infty ,a):a \in \mathbb{R}} \right\}$.
The derived set of $\{0\}$ is the set $(0,\infty)$ which is not closed because $0$ is limit point of that set.

7. Originally Posted by Plato
All of that is very true. But that is not the question.
The question is: Prove that the derived set of a set is closed. And that is not true.
Think of the space of real numbers, $\mathbb{R}$ with a topology generated by the collection $\left\{ {( - \infty ,a):a \in \mathbb{R}} \right\}$.
The derived set of $\{0\}$ is the set $(0,\infty)$ which is not closed because $0$ is limit point of that set.
Sorry! Question confusion.