TITLE: Accumulation points of sequences as limits of subsequences?
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: If you are interested in a counterexample, here is one:
For simplicity I assume that the sequence $s$ is just $(0,1,2,\dots)$ and $a\notin{\bf N}$ is some fresh element, we'd now like to construct a topology $\tau$ on $\{a\}\cup{\bf N}$ such that $(i)$ $a$ is limit point of $s$ and $(ii)$ no subsequence of $s$ converges to $a$.  I assume that $\tau$ is generated by $\{n\}$ for $n\in{\bf N}$ and $\{a\}\cup X$ for $X\in{\cal X}$ with some ${\cal X}\subseteq{\cal P}({\bf N})$ to be figured out yet.  If ${\cal X}$ is a filter then the neighborhoods of $a$ are exactly the sets $\{a\}\cup X$ for $X\in{\cal X}$, which I will also assume.
Condition $(i)$ demands that each $X\in{\cal X}$ is infinite.  Assume condition $(ii)$ were not satisfied, then some subsequence $s'$ of $s$ would have a tail living in $X$ for every $X\in{\cal X}$, i.e. if we call $X'$ the result of adding to $X$ the finitely many elements of $s'$ not in $X$ then $s'$ lives completely in $A:=\cap_{X\in{\cal X}} X'$.
Take ${\cal X}$ as some free ultrafilter, where free just means empty intersection.  Then all sets in ${\cal X}$ are infinite.  Since $A$ is infinite, we can just remove countably many elements but also leave countably many in place, the ultrafilter property ensures that this new set or its complement is contained in ${\cal X}$, in any case some set $B$ which lacks countably many elements of $A$.  But our construction tells us that $A$ should be contained in $B'$, a contradiction hence condition $(ii)$ is also satisfied.