I'm reading Abbott's Understanding Analysis (2nd edition). And in a chapter regarding closed sets, I'm faced with a theorem about limit points, which says
Theorem. A point $x$ is a limit point of a set $A$ if and only if $x = \lim{a_n}$ for some sequence $(a_n)$ contained in $A$ satisfying $a_n \neq x$ for all $n \in \mathbb{N}$.
And this is the definition of a limit point:
Definition. A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_\epsilon(x)$ of $x$ intersects the set $A$ at some point other than $x$.
But what if I let A be a set of real numbers $x$, satisfying $0 \leq x \leq 2$, and let $(a_n)$ be a sequence such that $a_n = 1 – 1/n$, so that $a_n \in A$ and $a_n \neq 1$ for all $n \in \mathbb{N}$, but $\lim{a_n} = 1$? By the definition of a limit point, I see that if $\epsilon \leq 1$, the $\epsilon$-neighborhoods of $x$ do not intersect the set $A$, which implies this $x$ is not a limit point of $A$.
Where are the points I am missing and wrong?
Best Answer
Let's call the first notion a "sequential limit point", and the second one just "limit point". The second one, that every open neighbourhood of $x$ intersects $A$ in a point different from $x$, is the one used in a general topology setting too (in metric spaces this can be specialised to $\varepsilon$-balls around $x$, as is done here). It's always true that if $x$ is a sequential limit point of $A$, then it is a limit point of $A$. This is easy to see, as any open ($\varepsilon$-)neighbourhood of $x$ contains infinitely many terms of the sequence and so plenty of points of $A$ different from $x$.
The reverse is also true in metric spaces (or more generally sequential spaces, but I'll restrict myself to metric here): if $x$ is a limit point of $A$ for each $n$ we can pick $a_n \in A \cap B(x,\frac{1}{n})$ with $a_n \neq x$. Then $a_n \to a$ (If $\varepsilon >0$ is given find $N$ such that $\frac{1}{N} < \varepsilon$ (or equivalently $N > \frac{1}{\varepsilon}$) and then for all $n \ge N$, $a_n \in B(a, \frac{1}{n})$ so $d(a_n, a) < \frac{1}{n} \le \frac{1}{N} < \varepsilon$, as required.) Hence $x$ is then also a sequential limit point of $A$.
I frankly don't see the point of your example: $a_n = 1-\frac{1}{n}$ are all in $A = [0,2]$ and converges to $1$. But $1$ is indeed a limit point of $A$ in both senses, in fact all points of $[0,2]$ are.