Limit and interior points of a set A general definition of limit point is the following:

A point $z_0$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_{\epsilon}(z_0)$ of $z_0$ intersects the set $A$ at some point other than $z_0$. 

Now a definition $\epsilon$-neighborhood:

An $\epsilon$-neighborhood of point $z_0$, denoted $V_{\epsilon}(z_0)$, is the set of all points $z$ for which $|z-z_0|<\epsilon$, for some $\epsilon$. That is, $V_{\epsilon}(z_0) = \{z \in \mathbb R\ : |z-z_0| < \epsilon\}$

Now consider the set $A$ formed by elements of the sequence $z_n=\frac{1}{n}$ for $n \in \mathbb N$, which has limit point $z=0$.
$$A = \{\frac{1}{n} : n \in \mathbb N\}$$
Why $z=1$ could not be a limit point of $A$? Does it mean that a limit point does not belong to the set itself? Is there another definition of interior point so that its neighborhood contains only points in $A$. In that case, which are the interior points of $A$?
Thanks in advance 
 A: The point $1$ is not a limit point of the set, because there is a neighbourhood of $1$ such that the only point in the set in that neighbourhood is $1$. Use, for example, the interval $(0.9,1.1)$.
In fact no point in the set is a limit point of the set. Around the point $\frac{1}{n}$, you can put the neighbourhood $(\frac{1}{n}-\frac{1}{4^n},\frac{1}{n}+\frac{1}{4^n})$ that contains no point of our set other than $\frac{1}{n}$.
To show that $0$ is a limit point of our set, take an open interval $(-\epsilon,\epsilon)$ about $0$, where $\epsilon$ is small (possibly very small) positive. There is an integer $N$ such that $\frac{1}{N}\lt \epsilon$. Then every point $\frac{1}{n}$ in our set with $n\ge N$ is in that neighbourhood.
So every open neighbourhood of $0$ contains a point of our set, indeed infinitely many points of our set.
Much more informally, we can get arbitrarily close to $0$ from within our set. 
A limit point of a set may or may not belong to the set. For example, let $S=(0,1)$, that is, all real numbers $x$ such that $0\lt x\lt 1$.  The number $1$ is a limit point of $S$, and is not in $S$.
Let $T=[0,1]$. The point $1$ is a limit point of $T$ and is in $T$. 
A: For a limit point $x$ of a set $S$, every neighborhood of $x$ contains infinitely many points of $S$.  In your example, not every neighborhood of 1 contains infinitely many points of the set.  For example, the open ball centered at 1 with radius 1/4 contains only 1.
A: We consider the set $E=\{1,\frac12,\frac13,\ldots\}$ in $\mathbb R$. No point of $E$ is an interior point of $E$. $1$ is not a limit point of $E$ because its $\frac12$-neighbourhood does not contain any point of the set other than itself. But $0$ is a limit point of $E$ because for every $\epsilon >0$, there is a positive integer $k$ so that $\frac1k <\epsilon$, and then $\frac1k$ is a point in $E$ that is not $0$ itself ($\frac1k$ is in its $\epsilon$-neighbourhood).
A limit point of a set may or may not belong to the set. An example of a limit point of a set which belongs to the set is the point $1$ which is a limit point of $[0,1]$ in $\mathbb R$. An example of a limit point of a set which does not belong to the set is the point $1$ which is a limit point of the set $(0,1)$ in $\mathbb R$.
