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Say I have a metric space $(X, d)$ and a set $A\in X$. I want to prove that if $a \in \overline{A}$ then there is a convergent sequence $\{x_n\}$ that converges to $a$. Could I have any help?

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Consider a sequence $(B_n)$ of open balls, centered at $a$, whose radii tend to $0$. – David Mitra Sep 2 '12 at 14:12
That is kind of the definition of a closure.. – Karolis Juodelė Sep 2 '12 at 14:12
up vote 2 down vote accepted

If $a \in \overline{A}$ (the closure of A), then there are two possibilities:

  1. $a \in A$ and it's an isolated point of $A$.
  2. $a$ is a limit point of $A$.

For the first case, consider the sequence whose terms are all $a$. Since $a \in A$, this sequence is in $A$. It's trivial to show that this sequence converges to $a$.

For the second case, consider a collection of neighborhoods of $a$ of the form: $B(a; \frac{1}{n})$ where $n \in \mathbb{N}$. Each element of this collection is an open ball centered at $a$ whose radius is $\frac{1}{n}$. Since $a$ is a limit point of $A$, for each neighborhood $B(a; \frac{1}{n})$ we can pick $x_n \in A$ for which $x_n \in B(a; \frac{1}{n})$. It's easy to show that $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $A$ that converges to $a$.

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This depends on the definition of the closure you're using. In the case of metric space, you can actually define the closure $\bar{A}$ as the set of limits of sequences of elements of $A$.

If the definition you are using is the general definition (which works for topological spaces too), that is the set of points of which any neighbourhood meets $A$, then the proof that the two formulations are equivalent is the following :

First, suppose $x$ checks the first definition, and let $V$ be a neighbourhood of $A$, and $\epsilon$ such that $B(x,\epsilon) \subset V$. Since $x$ is the limit of a sequence $(x_n)_n$ in $A$, we may find a $N$ such that $x_n \in B(x,\epsilon) \subset V$ for all $n\geq N$.

Conversely, suppose that $x$ checks the second definition. Then chose $x_n \in A$ such that $x_n \in B(x, \frac{1}{n})$, it is then obvious that $(x_n)_n$ converges to $x$.

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