Expressing Sequence as a set Let $X$ be a set and let $(x_n)$ be a sequence in $X$. Then obviously $\{x_n:~n\in \mathbb{N}\}$ is a subset of $X$. Is that set necessarily infinite? For example if the sequence is an eventually constant sequence, i.e., $x_n=a$ for all $n\geq n_0$ for some $n_0\in \mathbb{N}$, then is not the set $\{x_n:~n\in \mathbb{N}\}$ a finite subset of $X$.
On the other hand, if $\{x_n:~n\in \mathbb{N}\}$ is a finite subset of $X$, then is it always true that the sequence is eventually constant? In particular, what can we say about the sequence $(x_n)$, if $\{x_n:~n\in \mathbb{N}\}$ is a finite subset of $X$?
Edit: Let $(x_n)$ be a sequence such that $ \{x_n:~n\in \mathbb{N}\}$ is a finite subset of $X$, then $(x_n)$ contains an eventually contant subsequence.
Let $\{x_n:~n\in \mathbb{N}\}=\{y_1,y_2, \dots, y_k\}$. Consider an Equivalence relation on $\mathbb{N}$ by
$$p\sim q \iff x_p = x_q.$$
Then $\sim$ creates a partition on $\mathbb{N}$, say $A_\alpha$, where $\alpha$ runs over all disjoint equivalence classes. Obviously,
$$\mathbb{N}=\bigcup_{\alpha} A_\alpha.$$
We note that $\frac{N}{\sim}$ is finite. Otherwise, by choice axiom choosing exactly one member from each disjoint equivalence classes, we have $\{x_n:~n\in \mathbb{N}\}$ is infinite, a contradiction. Now index the disjoint equivalence classes by $J_1, J_2, \dots , J_k$ where $\{x_n:~n\in J_m\}=\{y_m\}$; $1\leq m \leq k$.
It is straightforward to see that at least one of $J_m$ is infinite, call it $\widetilde{J}$. Thereafter $\{x_n:~n\in \widetilde{J}\}$ is a constant subsequence of $(x_n)$.
 A: This possibly stems from a misunderstanding of the notation
$$
\{x_n:n\in\mathbb{N}\}
$$
where $(x_n)$ is a sequence in $X$ (call it the support of the sequence).
When $f\colon A\to B$ is a function, it's common to use
$$
\{f(a):a\in A\}
$$
which the above is a special case of, with $A=\mathbb{N}$, $B=X$ and $f(n)=x_n$. However this is an “abuse of notation” and it should be considered as a shorthand for
$$
\{b\in B:\exists\,a\in A(f(a)=b)\}
$$
Variations on the set builder notation are possible, but the general idea is always this.
You cannot recover the sequence from its support except in one trivial case, namely when the support is a singleton. For instance, if $p$ and $q$ are distinct points in $X$, there are sequences having $\{p,q\}$ as support that converge and sequences that don't.
Consider
$$
x_n=\begin{cases} p & n=0 \\ q & n>0 \end{cases}
\qquad\text{and}\qquad
y_n=\begin{cases} p & n \text{ even} \\ q & n \text{ odd} \end{cases}
$$
The sequence $(x_n)$ converges, the sequence $(y_n)$ doesn't.
Note also that the second sequence has subsequences that aren't eventually constant: if you remove the terms where $n\equiv2\pmod{4}$ or $n\equiv3\pmod{4}$, you obtain the same sequence as a subsequence: basically it is
$$
y_0=p,y_1=q,y_4=p,y_5=q,y_8=p,y_9=q,\dotsc
$$
On the other hand, if the support of $(x_n)$ is finite, there certainly exists a constant subsequence; let $\{a_1,\dots,a_r\}$ be the support and consider $A_i=\{n\in\mathbb{N}:x_n=a_i\}$ for $i=1,2,\dots,r$. One of the sets $A_i$ must be infinite and this will provide the required subsequence.
A: Let me redefine your question(s) first.
Let $X$ be a topological space, let $\left( x_n \right)$ be a sequence in $X$ and let $A=\{x_n\in X\vert x_n \text{ is a term of the sequence} \left(x_n \right)\}$ be the set corresponding to that sequence.

*

*Does the set $A$ have to be infinite?

*If the set $A$ can be finite, what can we say about the sequence?

Let me answer to these questions now:

*

*When representing sets, it doesn't matter how much you write an element in the set, it counts only once. For example, $B=\{1,2\}$ and $C=\{1,2,2,1,1,1,2,1,1,2\}$ are just the same set $(B=C).$ Now it should be clear that the answer of your first question is negative (as you supposed to be), and your example with the constant sequence holds.


*If the set $A$ is finite, then the sequence must be constructed from a finite number of constant subsequences, i.e. there exist constant sequences $(y_1),(y_2),\ldots,(y_k)$ and $I_1,I_2,\ldots ,I_k$ a partition of $\mathbb{N}$ such that $$x_n=\begin{cases}y_1 & \text{when } n\in I_1 \\ y_2 & \text{when } n\in I_2 \\ \vdots \\ y_k &\text{when } n\in I_k \end{cases}.$$
As a counter example that the sequence needs not to be a constant sequence consider $$x_n=\begin{cases} 0 & \text{if $n$ is odd} \\ 1 & \text{if $n$ is even}  \end{cases}.$$
A: For the counterexamples to what you ask, consider two distinct points $a,b\in X$ and the map \begin{align}&x_\bullet:\Bbb N\to X\\ &x_n=\begin{cases}a&\text{if }2\mid n\\ b&\text{if }2\nmid n\end{cases}\end{align}
It's hard to say what you can say of these sequences. A bunch of obvious things, I guess, but I wouldn't know how to give an exhaustive or interesting list of facts. Just keep in mind that $x_\bullet$ is a function with domain $\Bbb N$ and finite range, and go from there.
For the record, you should be aware of the fact that sequences may parameterize subsets of the space, but they aren't subsets themselves: sequences are functions from $\Bbb N$ (or something of sort) to the space of interest.
