I'm trying to extract more useful information from the following definition:
A point $x^*$ is an accumulation point of the sequence $\{x_n\}$ if, for every open set containing x , there are infinitely many indices such that the corresponding elements of the sequence belong to the open set.
I would like to say that this is equivalent to saying the following:
$x^*$ is an accumulation point if there exists a subsequence $x_{a_n}$ such that $lim_{n\rightarrow\infty} x_{a_n}=x^*$.
Is this true?
Best Answer
This is not true in general. If a space is first countable, i.e. has a countable basis at each point, then this result is true. Metric spaces are first countable.
To prove this, suppose $X$ is a first countable space and $A \subset X$. (In your case, $A = \{x_n : n \in \mathbb{N}$ where $(x_n)$ is a sequence.) Let $(U_n)_{n \in \mathbb{N}}$ denote a countable basis at the point $x^*$, where $x^*$ is an accumulation point of $A$. By definition, of accumulation point, there exists a $x_0 \in A$ and $x_0 \in U_0$. $U_0 \cap U_1$ is an open set. Again since $x^*$ is an accumulation point, there exists a $x_1 \in A$ and $x_1 \in U_0 \cap U_1$. Continue this to create a sequence $(x_n)_{n \in \mathbb{N}}$. Then since $U_n$ is a countable basis, you can prove that $\lim_{n \rightarrow \infty} x_n = x^*$. This is because given any open set $U$, there exists a $n$ such that $U_n \subset U$ such that $x^* \in U_n$ (definition countable basis). By construction, for all $m > n$, $x_m \in U_n \subset U$.
In metric spaces, the proof is bit nicer. You can take your countable basis to be $U_n = B_{\frac{1}{n}}(x^*)$. So using the definition of accumulation point, just choose $x_n$ such that $x_n \in B_\frac{1}{n}(x^*)$.