Constructing a sequence that converges to an accumulation point of a set My question is about characterizing accumulation points in terms of convergent sequences in a set $X$. I take the following definition of an accumulation point:


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*A point $x_0$ is an accumulation point of a set $X$ if all neighborhoods of $x_0$ contain an infinite points of $X$.


Consider the following assertion: $x_0$ is an accumulation point of $X$ if and only if there is a sequence $\{x_n\}_{n=1}^\infty$ in $X$ with $x_n \rightarrow x_0$ and $x_n \neq x_0$ for $n=1,2,...$
Proof: Take $\epsilon_1=1$. The existence of a point $x_1 \in X$ with $0<|x_1-x_0|<\epsilon_1=1$ is guaranteed. Now take $\epsilon_2=|x_1-x_0|/2$ and choose an $x_2 \in X$ with $0<|x_2-x_0|< \epsilon_2$. In general, choose $x_n \in X$ that satisfies $0<|x_n-x_0|< \epsilon_n=\frac{|x_1-x_0|}{2^n}<\frac 1{2^n}$. It is clear that $x_n \rightarrow x_0$ and $x_n \neq x_0$ for $n=1,2,...$
Question: Is this correct?
 A: If the space is metric or metrizable then your proof is correct. It also proves the equivalent statement:

A point $x_0$ is an accumulation point of a set $X$ if all neighborhoods of $x_0$ contain at least one point from $X$ other than $x_0$.

On the other hand, you cannot use the same argument for the arbitrary topological space. Clearly, existence of convergence sequence ensures that the point is an accumulation point.
For the first countable spaces you can apply this argument: consider nested neighborhood basis at $x_0$ which we denote $V_1,...,V_n,...$ and pick up any $x'_i\in V_i$, then $\lim\limits_{i\to\infty}x'_i = x_0$ since in neighborhood $U$ of $x_0$ contain all $V_i$ starting with some $i$.
Nested neighborhood basis $(V_i)_{i=1}^\infty$ is a neighborhood basis such that $V_{i+1}\subseteq V_i$. You can construct from the usual basis $B_1,...,B_n,...$ by taking $V_i = \bigcap\limits_{j=1}^i B_j$.
Unfortunately I cannot tell you if there exists a convergent sequence to an accumulation point for the spaces which are not first countable.
