# Thread: Closed set

1. ## Closed set

Hey, I was just wondering if anyone could tell me if this was about right.

Let E' be the set of limit points of E. Then E' is closed.

Proof:
Take an arbitrary limit point of E', say y.

Then, for any real $\displaystyle \epsilon$ > 0, there exists a point p in E' such that $\displaystyle p \in N_{\epsilon}(y)$.
But, for any p $\displaystyle \in$ E', there exists a real $\displaystyle \delta$, with $\displaystyle 0 < \delta < \epsilon$ such that, for some point q in E, $\displaystyle q \in N_{\delta}(p)$

But then, $\displaystyle q \in N_{\epsilon}(y)$ too, so y must be a limit point of E', and E' is closed by definition.

2. Originally Posted by Math Major
Hey, I was just wondering if anyone could tell me if this was about right.

Let E' be the set of limit points of E. Then E' is closed.

Proof:
Take an arbitrary limit point of E', say y.

Then, for any real $\displaystyle \epsilon$ > 0, there exists a point p in E' such that $\displaystyle p \in N_{\epsilon}(y)$.
But, for any p $\displaystyle \in$ E', there exists a real $\displaystyle \delta$, with $\displaystyle 0 < \delta < \epsilon$ such that, for some point q in E, $\displaystyle q \in N_{\delta}(p)$

But then, $\displaystyle q \in N_{\epsilon}(y)$ too, so y must be a limit point of E', and E' is closed by definition.
In a metric space this is the general idea.
But you need to take care: $\displaystyle \delta=\min\left\{d(p,y),\epsilon-d(p,y)\right\}$.
The insures subsetness.