# Thread: Limit points in a metric space

1. ## Limit points in a metric space

Let (X,d) be a metric space and assme that A$\displaystyle \subset$X. Prove that $\displaystyle x \in Cl(A)$ iff there exists a sequence in A converging to x.

2. Originally Posted by Andreamet
Let (X,d) be a metric space and assume that A$\displaystyle \subset$X. Prove that $\displaystyle x \in Cl(A)$ iff there exists a sequence in A converging to x.
Here's a rushed proof. you can streamline it.

Assume there is a sequence of points in $\displaystyle A$ that converges to $\displaystyle x$. Then $\displaystyle x$ lies in any closed set containing $\displaystyle A$. That is, $\displaystyle x \in \text{Cl}(A)$.

Conversely, assume $\displaystyle x \in \text{Cl}(A)$. Then the "ball" of radius $\displaystyle 1/k$ centered at $\displaystyle x$ will contain points in $\displaystyle A$. For each $\displaystyle k$, pick one of these points and call it $\displaystyle a_k$, then the sequence $\displaystyle \{ a_k \}$ converges to $\displaystyle x$

I assume you know how to describe a "ball" in a metric space. we derive this from the metric

3. Originally Posted by Jhevon
For each $\displaystyle k$, pick one of these points and call it $\displaystyle a_k$, then the sequence $\displaystyle \{ a_k \}$ converges to $\displaystyle x$

What theorem or axiom in mathematics will allow you to do this picking process and so be able to form the sequence in question??

4. Originally Posted by xalk
What theorem or axiom in mathematics will allow you to do this picking process and so be able to form the sequence in question??
gee, i don't know of a theorem that says we can do this. it just makes sense that we can based on our definitions.

what does it mean to be in $\displaystyle \text{Cl}(A)$? Well, it means that if $\displaystyle x \in \text{Cl}(A)$ then any "open ball" centered at $\displaystyle x$ will contain points in $\displaystyle A$. what does "open ball" mean in a metric space? the "open ball" centered at x is the set $\displaystyle \{ y \in A \mid d(x,y) < r , \text{ for some given } r > 0 \}$. so being in the closure means that this set is nonempty. if the set is nonempty, what's stopping us from picking elements in the set?

5. Originally Posted by Jhevon
gee, i don't know of a theorem that says we can do this. it just makes sense that we can based on our definitions.

what does it mean to be in $\displaystyle \text{Cl}(A)$? Well, it means that if $\displaystyle x \in \text{Cl}(A)$ then any "open ball" centered at $\displaystyle x$ will contain points in $\displaystyle A$. what does "open ball" mean in a metric space? the "open ball" centered at x is the set $\displaystyle \{ y \in A \mid d(x,y) < r , \text{ for some given } r > 0 \}$. so being in the closure means that this set is nonempty. if the set is nonempty, what's stopping us from picking elements in the set?