Convergence in a Topological Space I am trying to figure out how to prove the following problem from my topology homework:

"Let $X$ have the discrete topology and let $(a_n) \rightarrow b$. Prove that the sequence must be eventually constant; that is, the sequence must be of the form $( a_1, a_2, \dots , a_k, b, b, b, \dots)$."

I am aware that a convergence is defined as 

A sequence $x_1,x_2, \dots $ of points in a space $X$ converges to a point $x \in X$, if for each neighborhood $U$ of $x$, there exists an $N \in \mathbb{N}$ such that $\forall n \geq N :~ x_n \in U$ 

but I have no idea on an intuitive level how this defines convergence on a topological space. 
It seems to me by definition of convergence, you look at every open neighborhood of the convergence point. Doesn't that mean that it always converges? I can't seem to think of a concrete example of a sequence in a topological space not converging. 
Also, what I am further confused by is how my homework problem has $(a_n) \rightarrow b$ as given. Since it is given, isn't it eventually constant to $b$ by that very notion? 
I really would appreciate any help through hints, concrete examples, and clarifications, etc. as I have been stuck on this problem for quite a while now. 
Thanks!
 A: Let $X$ be a topological space, $\langle x_n:n\in\Bbb N\rangle$ a sequence of points of $X$, and $x\in X$. 

Definition $1$. $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ if and only if for each open nbhd $U$ of $x$ there is an $m\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m_U$. 

We sometimes express this informally by saying that every open nbhd of $x$ contains a tail of the sequence, meaning all of the terms from some point on.
This is a generalization of the familiar definition of convergence of sequences of real numbers: 

Definition $2$. $\langle x_n:n\in\Bbb N\rangle\to x$ if and only if for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $x_n\in(x-\epsilon,x+\epsilon)$ whenever $n\ge m_\epsilon$.

The sets $(x-\epsilon,x+\epsilon)$ for positive $\epsilon$ aren’t the only open nbhds of $x$ in the usual topology on $\Bbb R$, but it turns out that if $U$ is an open nbhd of $x$ in the usual topology on $\Bbb R$, then there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq U$, and we can use this fact to show that a sequence of real numbers converges by one definition if and only if it converges by the other. Thus, you know lots of examples of non-convergent sequences in $\Bbb R$ with its usual topology; $x_n=(-1)^n$ and $x_n=n$ give you two of them.
Here’s another fairly simple example. Let $X=\Bbb R$, and let 
$$\tau=\{U\subseteq X:X\setminus U\text{ is a countable set}\}\cup\{\varnothing\}\;.$$
Check that $\langle X,\tau\rangle$ is a topological space. (This topology $\tau$ is called the co-countable topology on $X$.) Let $\langle x_n:n\in\Bbb N\rangle$ be any sequence of distinct points of $X$, and let $x$ be any point of $X$; I claim that $\langle x_n:n\in\Bbb N\rangle$ does not converge to $x$. To see this, let $S=\{x_n:n\in\Bbb N\}\setminus\{x\}$, and let $U=X\setminus S$. Then $X\setminus U=S$, which is countable, so $U\in\tau$, i.e., $U$ is an open set in this space. Moreover, $x\notin S$, so $x\in U$, and $U$ is therefore an open nbhd of $x$. Clearly no $x_n$ belongs to the set $U$, so there cannot possibly be an $m_U\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m_U$, and therefore $\langle x_n:n\in\Bbb N\rangle$ does not converge to $x$. (In fact it turns out that this is another space in which the only convergent sequences are the ones that are constant from some point on; you might try to prove that.)
In your specific problem the topology of $X$ is the discrete topology, meaning that every subset of $X$ is an open set, and you’re given that the sequence $\langle a_n:n\in\Bbb N\rangle$ converges to $b$ in $X$. Since every subset of $X$ is open, $\{b\}$ is an open nbhd of $b$; and since $\langle a_n:n\in\Bbb N\rangle$ converges to $b$, there must be some $m\in\Bbb N$ such that $a_n\in\{b\}$ whenever $n\ge m$. If $a_n\in\{b\}$, what is $a_n$?
A: Sequences in general need not converge. Consider $a_n=n$ for $n\in\Bbb N$ as a sequence of reals (or naturals for that matter) in the usual topology.
Convergent sequences in general need not eventually be constant. Consider $b_n=\frac1{n+1}$ for $n\in\Bbb N$ as a sequence of reals--this is never constant, but does converge (to what?).
On the other hand, in any space, sequences that are eventually constant do converge.
In discrete spaces in particular, convergent sequences must be eventually constant. To see this, note that if $x_n\to x$ in a discrete space, then $\{x\}$ is an open neighborhood of $x$. What then can you conclude about the sequence and the set $\{x\}$ by definition of convergence?
