The limit points of an open interval (open set) I am not sure I understand why for the open set $(a,b)$, the limit points are $[a,b]$. Why are $a,b$ now included as limit points? Is this because we can somehow find a sequence in the open interval converging to these points? 
 A: Indeed we can find sequences that converge to those points. 
Let $d = b-a$ and $\epsilon_n = \frac{d}{n}$. For each $n \in \mathbb{N}$ choose a point $x_n \in (a, a+\epsilon_n)$. Now we claim that that the sequence $\{x_n\}$ converges to $a$. 
Let $\epsilon > 0$ then by construction we have
$$|x_n - a| < \epsilon_n = \frac{d}{n}$$
Thus we can let $N\in \mathbb{N}$ with $N > \frac{d}{\epsilon}$ so that for all $n \ge N$ we have 
$$|x_n - a| < \epsilon_n = \frac{d}{n} < \epsilon$$
which means that $\lim x_n = a$. Thus $a$ is a limit point of $(a,b)$ and by similar approach (which I will leave to you), it is possible to show that $b$ is also a limit point.
$\boxtimes$
So what we have really done here is look at smaller and smaller intervals contained in $(a,b)$ that are close to $a$ and pick points from each of these intervals. These points form our sequence and the limit of this sequence is $a$. If you need help on constructing a sequence that converges to $b$, think about what intervals you would need to pick points from. 
A: So according to your definition of limit points, it sounds like the points in $\overline{A}$, if so, yes, because the points $x_{n}=a+\dfrac{b-a}{2n}$, $n=1,2,...$ belong to $(a,b)$ and are such that $x_{n}\rightarrow a$.
A: $a$ is a limit point
Sequence $x_n=a+\frac{1}{n} \in (a,b)\forall n \in \mathbb N(\because$ consequence of archemedean property). We know that $x_n \to a$ as $n \to \infty$. Hence $a$ is the limit point of $(a,b)$. ($\because$ $x\in \mathbb{R}$ is a limit point of $A$ iff there is a sequence of points $\{x_{n}\}\subset A$ which are different from $x$ and converges to $x$). Similar argument for $b$.
A: We can also try to find the closure $C$ of $(a,b)$. By definition $C$ is the intersection of all closed sets containing $(a,b)$ and $C$ will contain all limit points of $(a,b)$. The closed sets containing $(a,b)$ are of the following two forms 
$$A=(-\infty,d]\cup[a,+\infty)\space\space(a>d)$$ and
$$B=(-\infty,b]\cup[c,+\infty)\space\space(b<c)$$ so
$$C=[a,b]$$
On the other hand, by definition $C$ is the union of $(a,b)$ and the derived set $D$ of $(a,b)$ which is the set of all limit points of $(a,b)$. But $a,b\notin(a,b)$, they must come from $D$. So $a,b$ are limit points.
A: Suppose an open interval $(c, d)$ such that it contains only $a$, we can verify it is open in Euclidean topology (Topology Without Tears, definition 2.1.1), $(c, d)$ has no intersection with $(a, b)$, so, $a$ is not a limit point of $(a, b)$. Maybe I am wrong.
