I understand in your comment above to Jonas' answer that you would like these things to be broken down into simpler terms.
Think about limit points visually. Suppose you have a point $p$ that is a limit point of a set $E$. What does this mean? In plain terms (sans quantifiers) this means no matter what ball you draw about $p$, that ball will always contain a point of $E$ different from $p.$
For example, look at Jonas' first example above. What you should do wherever you are now is draw the number line, the point $0$, and then points of the set that Jonas described above. Namely draw $1, 1/2, 1/3,$ etc (of course it would not be possible to draw all of them!!).
Now an open ball in the metric space $\mathbb{R}$ with the usual Euclidean metric is just an open interval of the form $(-a,a)$ where $a\in \mathbb{R}$. Now we claim that $0$ is a limit point. How?
Given me an open interval about $0$. For now let it be $(-0.5343, 0.5343)$, a random interval I plucked out of the air. The question now is does this interval contain a point $p$ of the set $\{\frac{1}{n}\}_{n=1}^{\infty}$ different from $0$? Well sure, because by the archimedean property of the reals given any $\epsilon > 0$, we can find $n \in N$ such that
$$0 < \frac{1}{n} < \epsilon.$$
In fact you should be able to see from this immediately that whether or not I picked the open interval $(-0.5343,0.5343)$, $(-\sqrt{2},\sqrt{2})$ or any open interval.
Now let us look at the set $\mathbb{Z}$ as a subset of the reals. What you do now is get a paper, draw the number line and draw some dots on there to represent the integers. Can you see why you are able to draw a ball around an integer that does not contain any other integer?
Having understood this, looks at the following definition below:
$\textbf{Definition:}$ Let $E \subset X$ a metric space. We say that $p$ is a limit point of $E$ if for all $\epsilon > 0$, $B_{\epsilon} (p)$ contains a point of $E$ different from $p$.
$\textbf{The negation:} $ A point $p$ is not a limit point of $E$ if there exists some $\epsilon > 0$ such that $B_{\epsilon} (p)$ contains no point of $E$ different from $p$.
From the negation above, can you see now why every point of $\mathbb{Z}$ satisfies the negation? You already know that you are able to draw a ball around an integer that does not contain any other integer.
You can partition your set E into $E_{2n}\cup E_{2n+1}$, and then you can see that
$E_{2n}=\frac {n}{n+1}$ , which is strictly increasing, and $E_{2n+1}= \frac{1}{n(n+1)}$, which is strictly decreasing. This means that the limit points of E will be the LUB of $E_{2n}$ and the INF of $E_{2n+1}$ , which are $1$ and $0$ respectively, as you correctly pointed out (the only limit point of a bounded non-decreasing sequence of Reals is its LUB, a.k.a SUP, and the only limit point of a non-increasing sequence of Reals is its GLB a.k.a INF). This means that beyond a certain N, all your terms $E_{2n}$ will be within $e $ of 1, for any $e>0$, and all the odd terms $E_{2n+1}$ will be within $e'>0$ of $0$. This makes any possible third limit point $p$ impossible, as this potential third limit point cannot be approached within less than neither $e< \frac{1-p}{2}$ by points approaching 1, nor can it be approached closer than $e'<\frac{p}{2}$ by the terms approaching $0$, by choosing $N,N'$ large-enough for $E_{2n},E_{2n+1}$ respectively
Best Answer
There is no necessary relationship between the two sets.
Let $A$ be a set in a topological space $X$. A point $x$ is in the interior of $A$ if there is an open set $U$ such that $x\in U\subseteq A$; in particular this implies that $x\in A$. A point $x$ is an accumulation point of $A$ if for each open set $U$ containing $x$, $U\cap(A\setminus\{x\}\ne\varnothing$; in words, if every open nbhd of $x$ contains at least one point of $A$ different from $x$. This does not imply that $x\in A$.
Take the space $\Bbb R$ with the usual topology as a familiar example. The set $\Bbb Z$ has empty interior: for any $n\in\Bbb Z$, no matter how small an $\epsilon>0$ you take, $(n-\epsilon,n+\epsilon)\nsubseteq\Bbb Z$, so $n$ is not in the interior of $\Bbb Z$. $\Bbb Z$ also has no accumulation points: if $x\in\Bbb R\setminus Z$, there is an integer $n$ such that $n<x<n+1$, and $(n,n+1)$ is then an open nbhd of $n$ that contains no point of $\Bbb Z$; and if $n\in\Bbb Z$, $(n-1,n+1)$ is an open nbhd of $n$ that contains no point of $\Bbb Z\setminus\{n\}$.
The set $\left\{\frac1n:n\in\Bbb Z^+\right\}$, on the other hand, has empty interior and exactly one accumulation point, $0$. In this case the interior of the set is its set of accumulation points. But now consider the following sets:
$\Bbb Q$ also has empty interior, and every real number is an accumulation point of $\Bbb Q$.
(In each case you should try to prove the assertions.)
If we look at $\Bbb Z$ as a space in its own right, with the discrete topology, then every subset of $\Bbb Z$ is open, and no point of $\Bbb Z$ is an accumulation point of any subset of $\Bbb Z$. Thus, if $A\subseteq\Bbb Z$, then the interior of $A$ is $A$ itself, but $A$ has no accumulation points.