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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:

  • 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?

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Are we looking at real numbers? In general we have no norm, "$|\cdot|$", on a topological space. – AD. Oct 20 '11 at 8:01
I should have mentioned this. Yes we are looking at real numbers. – Iconoclast Oct 20 '11 at 11:01
Then it is correct! – AD. Oct 20 '11 at 11:07
@Iconoclast: Sorry then for confusing you with so general stuff - confused myself by the tag [general-topology] – Ilya Oct 20 '11 at 11:22
@Gortaur: No problem. On the contrary it was informative. I guess the tag i chose was a bit misleading. – Iconoclast Oct 20 '11 at 13:12
up vote 1 down vote accepted

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.

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You asked, whether there are spaces which are not first countable and still have this property. Yes, they are such spaces. You can learn about them if you search for Frechet-Urysohn spaces, see e.g. this answer:… – Martin Sleziak Oct 20 '11 at 11:26
@MartinSleziak: Thank you very much – Ilya Oct 20 '11 at 11:31

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