Let $(M, d)$ be a metric space.
Let $a_1 = 1$ and for $n \in \N$, define $a_{n+1} = \sqrt{3+2a_n}$.
Suppose that $(a_n)_n$ is a sequence in a metric space $(M, d)$ which converges to a limit $a$. Prove that $\set{a_n\mid n \in \N} \cup \set a$ is compact.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Suppose that $(a_n)_n$, $(b_n)_n$ are sequences in a metric space $(M, d)$ so that \[\limni d(a_n, b_n) = 0.\] Prove that both sequences converge to the same limit, or provide a counterexample.
A topological space is a pair $(X, \mathscr{T})$ where $X$ is a set, and $\mathscr{T}$ is a collection of subsets of $X$ so that $\emptyset, X \in \mathscr{T}$, and $\mathscr{T}$ is closed under finite intersections and arbitrary unions (that is to say, if $G_1, \ldots, G_n \in \mathscr{T}$ then $G_1\cap\cdots\cap G_n \in \mathscr{T}$, and if $(G_\alpha)_\alpha$ is a collection of elements of $\mathscr{T}$ then $\bigcup_\alpha G_\alpha \in \mathscr{T}$). The elements of $\mathscr{T}$ are said to be open, and a set $F \subseteq X$ is said to be closed if $F^c \in \mathscr{T}$. We refer to $\mathscr{T}$ as a topology on $X$.
Notice that if $(M, d)$ is a metric space, then $\mathscr{T}_d = \set{U \subseteq M \mid U \text{ is open with respect to } d}$ is a topology on $M$. We say that the topology $\mathscr{T}_d$ is induced by the metric $d$.
A topological space $(X, \mathscr{T})$ is said to be metrizable if there is a metric on $X$ which induces $\mathscr{T}$. Note that this metric need not be unique!
A sequence $(x_n)_n$ in a topological space $(X, \mathscr{T})$ is said to converge to $x \in X$ if for every open set $G \in \mathscr{T}$ with $x \in G$, there are only finitely many $n \in \N$ for which $x_n \notin G$. (Remember that we proved this was equivalent to our definition of convergence in metric spaces.)