# Limit points in a metric space

• Apr 5th 2009, 03:45 PM
Andreamet
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.
• Apr 5th 2009, 04:27 PM
Jhevon
Quote:

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
• Apr 6th 2009, 08:06 AM
xalk
Quote:

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??
• Apr 6th 2009, 06:56 PM
Jhevon
Quote:

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?
• Apr 7th 2009, 02:11 AM
xalk
Quote:

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?