How to understand cluster points? We talked about cluster points yesterday in class and I am now very confused. I know the formal definition of the limit: $a$ is a cluster point, if for every $\epsilon > 0$, there exist infinitely many natural numbers $n ∈ ℕ$ such that $|a_n -a|<\epsilon$
Okay. I think I sort of understand this intuitively. If we have some sort of open interval around $a$, then a is a cluster point because we can fit infinitely many elements from a set or sequence into that interval no matter how small it is ($\epsilon$ is). The integer set does not contain any cluster points because it every element is isolated. Okay cool.
But I still cannot do this:
$1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, ... .$
Shouldn't there be no cluster points? If I draw an open interval around 1, there are no other elements from the sequence except 1 that I can fit inside it since every elements of the sequence are integers.
Consider:

In this case, I think $1$ and $-1$ are the cluster points. I don't think $0$ can be a cluster point for that I can't construct a small interval to include elements from the sequence.
and what happens if we have a something like this

my teacher gave an explanation of the definiton of cluster point and that's it :(
 A: You're confusing the cluster points of a sequence with the accumulation points of a set. For the former, the $n$'s play a role! Say $(a_n)_{n\in \mathbb{N}}$ is the constant sequence $a_n=a_0$, then $a_0$ is a cluster point of the sequence.
Why? Well, let's check the definition. In order to prove that $a_0$ is a cluster point of the sequence, we let $\varepsilon>0$ be given and consider the set
$$
A_{\varepsilon}=\{n|\; |a_n-a_0|\leq \varepsilon\}
$$
and we're supposed to prove that this set is infinite.
However, clearly, $A_{\varepsilon}=\mathbb{N}$, which is obviously infinite, so $a_0$ would be a cluster point of the sequence. Similar things are true for the examples you listed. We do count repeat occurences of the same element in a sequence with multiplicity.
A: Some of what partially answers the question is given in the comments, and I might later try to expand what follows to include concerns discussed there. However, for now I will only discuss cluster points for the third sequence. I will show in great detail (for expository purposes) that the set of all cluster points of the third sequence is $\{-\frac{1}{2},\,0\}.$
For cluster points we don't care what $a_1$ is, so let's ignore $a_1.$ Letting $k=1,2,3,\ldots$ we get
$$a_2 = 0 \;\; \text{and} \;\; a_4 = -\frac{1}{4} \;\; \text{and} \;\; a_6 = -\frac{2}{6} \;\; \text{and} \;\; a_8 = -\frac{3}{8} \;\; \text{and} \;\; a_{10} = -\frac{4}{10} \;\; \text{and} \;\; \cdots $$
(fractions not reduced so that the pattern is easy to see), and we get
$$a_3 = \frac{\sin 3}{\sqrt 3} \;\; \text{and} \;\; a_5 = \frac{\sin 5}{\sqrt 5} \;\; \text{and} \;\; a_7 = \frac{\sin 7}{\sqrt 7} \;\; \text{and} \;\; a_9 = \frac{\sin 9}{\sqrt 9} \;\; \text{and} \;\; \cdots $$
It is clear that $-\frac{1}{2}$ is a cluster point, because given any open interval about $-\frac{1}{2},$ we have infinitely many (in fact, all but finitely many) of the even-numbered terms in that interval. The assertion about open intervals is because the limit of the even-numbered exists and is equal to $-\frac{1}{2}.$
It is clear that $0$ is a cluster point, since the limit of the odd-numbered terms exists and is equal to $0.$ The limit is $0$ because the odd-numbered terms are fractions with numerators between $-1$ and $1$ and denominators that become (and remain) arbitrarily large.
Therefore, $\{-\frac{1}{2},\,0\}$ is a SUBSET of the set of all cluster points of the third sequence. To show that $\{-\frac{1}{2},\,0\}$ is EQUAL to the set of all cluster points of the third sequence, I will show that no other real number is a cluster point of the third sequence.
To this end, let $x$ be a real number such that $x \neq -\frac{1}{2}$ and $x \neq 0.$ I will show that $x$ is NOT a cluster point of the third sequence. To show that a “for all $\epsilon > 0$ something” statement is false (note that the definition of “$x$ is a cluster point” is a “for all $\epsilon > 0$ something” statement), it suffices to show that the corresponding “there exists $\epsilon > 0$ not-something” statement is true. Specifically, I will show there exists $\epsilon > 0$ such that it is false that infinitely many terms of the sequence are a distance of at most $\epsilon$ from $x.$
Intuitive idea for what follows: Because the even-numbered terms converge to $-\frac{1}{2}$ and the odd-numbered terms converge to $0,$ it follows that the terms will get closer and closer to these two numbers (and no other numbers; I'll show this below), which implies that the terms will NOT get (and remain) arbitrarily close to $x,$ because the minimum of the distances of $x$ from these two numbers is positive. Thus, even we imagine $x$ to be very close to one of these two numbers, what happens for terms far enough out in the sequence is that each of the terms will be super-close to $-\frac{1}{2}$ or $0,$ so close in fact that $x$ will seem to be far off in the distance like a distant star compared to a person standing next to you on your left or a person standing next to you on your right. (In this analogy, the distant star is $x,$ the person who could be next to you is $-\frac{1}{2}$ or $0,$ and you is any term that is far enough out in the sequence.)
Let $d$ be the minimum of the distances of $x$ from $-\frac{1}{2}$ and $0.$ That is, let $d = \min \left\{\left|x - \left(-\frac{1}{2}\right)\right|,\;\left|x - 0\right|\right\}.$ Since $x$ is not equal to $-\frac{1}{2}$ or $0,$ it follows that $d > 0.$ Now let $\epsilon = \frac{d}{2}$ and note that $\epsilon > 0.$ Since the even-numbered terms converge to $-\frac{1}{2},$ there exists a positive integer $N_1$ such that at most $N_1$ many of the even-numbered terms are NOT less than a distance of $\epsilon$ from $-\frac{1}{2}.$ Since the odd-numbered terms converge to $0,$ there exists a positive integer $N_2$ such that at most $N_2$ many of the odd-numbered terms are NOT less than a distance of $\epsilon$ from $0.$ It follows that all but at most $N_1 + N_2$ many of the terms (whether even-numbered or odd-numbered) are a distance of less than $\epsilon$ from one of the two numbers $-\frac{1}{2}$ and $0.$
Therefore, all but at most $N_1 + N_2$ many of the terms are a distance of more than $\epsilon$ from $x,$ since anything a distance less than $\frac{d}{2}$ from one of the two numbers will be a distance of more than $\frac{d}{2}$ from $x.$ To see this geometrically, draw a number line, mark points for $-\frac{1}{2}$ and $0,$ draw two open intervals of radii $\frac{d}{2}$ whose centers are these two points (these two open intervals may overlap, but that won't matter), and notice that $x$ is a distance of more than $\frac{d}{2}$ from any point within either of these two intervals. To see this algebraically, let $z$ represent either $-\frac{1}{2}$ or $0,$ and let $y$ be a number less than a distance of $\frac{d}{2}$ from $z$ $(y$ represents an appropriate term of the sequence, and we have $|y - z| < \frac{d}{2}),$ and note that if we assume (for a later contradiction) $|x - y| \leq \frac{d}{2},$ then by the triangle inequality we have
$$|x - z| \; \leq \; |x - y| + |y - z| \; \leq \; \frac{d}{2} + |y - z| \; < \; \frac{d}{2} + \frac{d}{2} \; = \; d,$$
which contradicts the fact that $|x - z| \geq d$ (i.e. the distance from $x$ to either of the numbers $-\frac{1}{2}$ or $0$ is at least $d),$ and hence $|x - y| \leq \frac{d}{2}$ must be false, and thus we have $|x - y| > \frac{d}{2}$ (i.e. the sequence term $y$ is a distance of more than $\frac{d}{2}$ from $x).$
Thus, to repeat what the stuff just above shows, we know that all but at most $N_1 + N_2$ many of the terms are a distance of more than $\epsilon$ from $x.$ That is, only finitely many of the terms are a distance of less than $\epsilon$ from $x,$ and hence it is false that infinitely many of the terms are a distance of less than $\epsilon$ from $x,$ which implies that $x$ cannot be a cluster point of the third sequence.
Generalization: The argument above can be adapted to prove that if a sequence is a finite union of sequences, each of which converges, then the set of all cluster points of the original sequence is the set whose elements are the limits of those finitely many sequences.
