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.

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- Apr 5th 2009, 03:45 PMAndreametLimit 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 PMJhevon
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 AMxalk
- Apr 6th 2009, 06:56 PMJhevon
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 AMxalk
- Apr 7th 2009, 03:57 PMJhevon