Convergence in a Topological Space

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 $a_n$?