What do Steen and Seebach mean when discussing limit points of sequences in Particular Point Topology In "Counterexamples in Topology" by Steen and Seebach (2ed: 1978), they are discussing the Particular Point Topology, which they define as:
"On any set $X$, we can define the open sets of a topology to be $\varnothing$ and any subset of $X$ that contains a particular point $p$. We distinguish three cases, finite, countable, and uncountable according to the size of $X$."
(It is of course assumed that $p \in X$.)
Their first article goes:


*

*The sequences $\langle a_i \rangle$ which converge are those for which the $a_i \ne p$ are equal for all but a finite number of indices. The only accumulation points for sequences are the points $b_j$ that the $a_i$ equal for infinitely many indices. So any countably infinite set containing $p$ has a limit point, but never an accumulation point when considered as a sequence in any ordering.


This is how S&S define an accumulation point of a sequence:
"... every open set containing $p$ contains infinitely many terms of the sequence. In this case $p$ is called an accumulation point of the sequence."
That is, an accumulation point $\alpha$ is such that:
$\forall U \in \tau: x \in U \implies \{n \in \mathbb N: x_n \in U\}$ is infinite (where $\tau$ is the topology in question).
My understanding (or lack of it) is as follows.
By definition, a sequence converges to $\alpha$ iff there exist only a finite number of open sets containing $\alpha$ which do not contain any given term of $\langle a_i \rangle$.
But consider the sequence $\langle a_i \rangle = \left\langle{\dfrac 1 i}\right\rangle_{i \in \mathbb N}$ in the particular point space $(\mathbb R, \tau_p)$ where $p = 1$ and $\mathbb R$ denotes the reals.
$\langle a_i \rangle$ converges to $0$, which does not equal $1$, but none of the $a_i$ are equal. All of the sets of $(\mathbb R, \tau_p)$ of the form $\left[{0, \dfrac 1 n}\right] \cup \{1\}$ are open in the particular point topology, so I don't understand that first sentence of article $1$.
Hence there seems to be a convergent sequence for which $a_i \ne p$ are not equal for all but a finite number of indices. And so the sentence about accumulation points is equally questionable.
As for the rest of the article, I can't get my head round it until I resolve my problem with these first bits.
Any insight appreciated, and a full explanation of the whole will be greatly appreciated and gratefully received.
 A: You misunderstand the definition of convergence

By definition, a sequence converges to $\alpha$ iff there exist only a finite number of open sets containing $\alpha$ which do not contain any given term of $\langle a_i \rangle$.

No It's actually more similar to the definition of acumulation point that you gave :
$$\forall U \in \tau: \alpha \in U \implies \{n \in \Bbb N: x_n \in U\} \text{ only misses finitely many points of } \Bbb N$$
So all open neighbourhoods of the limit contain "almost all" (at most finitely many exceptions allowed) terms of the sequence.
So your sequence does not converge to $0$ in $\tau_p$: $\{0,p\}$ is a neighbourhood of $0$ that contains at most two terms of the sequence (as the sequence has all distinct terms).
The sequence $0,1,0,1, \ldots$ has at least accumulation points $0$ and $1$ (in any topology) and in $\tau_p ( p=1) $ it doesn't have any other, and no limit. This is in accordance with Steen and Seebach's statement. 
A: This is my attempt to answer my own question, which is "make sense of that first sentence of the quoted passage."
Let $\langle a_i \rangle$ be a convergent sequence in a particular point space $T = (S, \tau_p)$, where $p$ is the particular point and $\tau$ is the set of all subsets of $S$ which contain $p$.
Let $\langle a_i \rangle$ converge to $\alpha$.
By definition of convergent sequence, all open sets of $T$ contain all but finitely many terms of $\langle a_i \rangle$.
This includes $\{\alpha, p\}$.
So all but a finite number of terms of $\langle a_i \rangle$ is equal either to $\alpha$ or $p$.
Hence all but a finite number of terms of $\langle a_i \rangle$ such that $a_i \ne p$ is equal to $\alpha$.
Job done.
A: I'll try to give a full overview of convergence of sequences and accumulation points of sequences in $X$ with the particular point topology (with $p\in X$ as the "particular point").  So let $(a_n)_n$ be a sequence in $X$.
Limits of sequences
By definition, $(a_n)_n$ converges to a point $a$ if every nbhd of $a$ contains all $a_n$ for sufficiently large $n$.  In the case that there is a smallest nbhd $V$ of $a$ (which is the case for each point in the particular space topology), it is equivalent to require that all $a_n$ belong to $V$ for large enough $n$.  For $a=p$, that smallest nbhd is $\{p\}$.  For $a\ne p$, that smallest nbhd is $\{a,p\}$.  So we get:

*

*$(a_n)_n$ converges to $p$ exactly when $a_n=p$ for all sufficiently large $n$ (the sequence is eventually constant equal to $p$).


*$(a_n)_n$ converges to $a\ne p$ exactly when $a_n\in\{a,p\}$ for all sufficiently large $n$.
Accumulation points of sequences
By definition, the point $a\in X$ is an accumulation point of the sequence $(a_n)_n$ if every nbhd of $a$ contains $a_n$ for infinitely many indices $n$.  Similar to limits, it is enough to require this for the smallest nbhd of $a$ in the topology of $X$.  We get:

*

*$p$ is an accumulation point of $(a_n)_n$ exactly when $p$ occurs infinitely many times in the sequence.


*$a\ne p$ is an accumulation point of $(a_n)_n$ exactly when $a$ occurs infinitely many times in the sequence or $p$ occurs infinitely many times in the sequence.
In particular, if $p$ occurs infinitely many times, every point of $X$ is an accumulation point of the sequence.
Examples
(taking $a$, $b$ distinct points, both different from $p$)

*

*The sequence $(p,p,p,\dots)$ converges to all points of $X$.  All the points of $X$ are accumulation points of the sequence.

*The sequence $(a,a,a,\dots)$ converges to $a$ only, and $a$ is its only accumulation point.

*The sequence $(a,p,a,p,a,p,\dots)$ converges to $a$ only.  All the points of $X$ are accumulation points of the sequence.

*The sequence $(a,b,p,a,b,p,a,b,p,\dots)$ does not converge.  All the points of $X$ are accumulation points of the sequence.

*The sequence $(a,b,a,b,a,b,\dots)$ does not converge.  Its accumulation points are $a$ and $b$.

