# Thread: Is derived set closed in any topological space?

1. Maybe.
Hope I can find a counterexample that A' is not closed in the future. Thank you Plato, for your patient replies.

2. ## A subset of a T0 space whose derived set is not closed

Consider $\displaystyle \mathbb{R}$ with the topology generated by $\displaystyle \{(-\infty,a) \mid a \in \mathbb{R} \}$, which is a T0 space. The derived set of $\displaystyle \{0\}$ is $\displaystyle (0, \infty)$, which is open.

3. A beautiful counterexample! Thank you for your brilliant first-post.

4. Originally Posted by austinmohr
Consider $\displaystyle \mathbb{R}$ with the topology generated by $\displaystyle \{(-\infty,a) \mid a \in \mathbb{R} \}$, which is a T0 space. The derived set of $\displaystyle \{0\}$ is $\displaystyle (0, \infty)$, which is open.
Ack - a typo. The set $\displaystyle (0, \infty)$ is neither open nor closed in this topology. The counterexample still works, however.

5. Originally Posted by austinmohr
Consider $\displaystyle \mathbb{R}$ with the topology generated by $\displaystyle \{(-\infty,a) \mid a \in \mathbb{R} \}$, which is a T0 space. The derived set of $\displaystyle \{0\}$ is $\displaystyle (0, \infty)$, which is open.
Originally Posted by austinmohr
Ack - a typo. The set $\displaystyle (0, \infty)$ is neither open nor closed in this topology. The counterexample still works, however.
If this is Lynn Steen’s Left Order Topology, then isn’t it true that $\displaystyle 0$ is a limit point of $\displaystyle \{0\}?$
If so, then is $\displaystyle 0$ in the derived set of $\displaystyle \{0\}?$
If that is so, what is its complement?

6. Originally Posted by Plato
If this is Lynn Steen’s Left Order Topology, then isn’t it true that $\displaystyle 0$ is a limit point of $\displaystyle \{0\}?$
If so, then is $\displaystyle 0$ in the derived set of $\displaystyle \{0\}?$
If that is so, what is its complement?
We may be using slightly different definitions. I use "derived set" to mean the set of all accumulation points of a set $\displaystyle A$. Hence, if $\displaystyle A$ is singleton, it cannot be contained in its own derived set by definition.

7. Originally Posted by austinmohr
We may be using slightly different definitions. I use "derived set" to mean the set of all accumulation points of a set $\displaystyle A$. Hence, if $\displaystyle A$ is singleton, it cannot be contained in its own derived set by definition.
You are correct. I was not thinking of a singleton set.
In fact, I don't know what I had in mind.
Maybe $\displaystyle ( - \infty ,0]$ I really don't know.

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# show that the derived set of A is a closed set

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