Give an example of a topological space $X$ and a finite subset $A$ of $X$ such that $\bar{A}$ is an infinite set Give an example of a topological space $X$ and a finite subset $A$ of $X$ such that $\bar{A}$ is an infinite set.
I have thought a lot about this question and all I can think of is to consider $\mathbb{R}$ with the discrete topology. In this topology, the set $A=\left \{  1,2\right \}$ is open and $\overline{A}$ is infinite because all the sets have $1$ and $2$, which are infinite… Is this reasoning right? Many thanks.
 A: There is of course the trivial or indiscrete topology on any infinite set.
You can also cook up an example using a special point.
Let $X$ be an infinite set and $x_0\in X$. We aim to define a topology making every point a limit point of $\{x_0\}$. Define $\tau$ to be the collection of subsets of $X$ which contain $x_0$, with the empty set. Verify that this is a topology on $X$. (In fact, its even closed under arbitrary intersections.)
Then $\overline{\{x_0\}}=X$. Indeed, if $x\in X$, then every open set $U$ containing $x$ must have $x_0\in U$ and hence $U\cap\{x_0\}\ne\emptyset$. Therefore $x\in\overline{\{x_0\}}$, which proves the closure $\overline{\{x_0\}}=X$ of the finite set $\{x_0\}$ is infinte.
A: Take $(\mathbb{R},\tau)$, with $\tau=\{\emptyset,\mathbb{R}\}\cup\{(-\infty,a)\,|\,a\in\mathbb R\}$. Then $(\forall x\in\mathbb{R}):\overline{\{x\}}=[x,+\infty)$. More generally, if $F\subset\mathbb R$ is finite and not empty, then $\overline F=[\min F,+\infty)$.
A: Let $S$ be an infinite set such that $\mathbb{R}\cap S = \emptyset$. Consider the topological space $\mathbb{R} \cup S$, where the open sets are of the form 
$$\begin{cases}
U,  & \text{if $0\notin U$} \\
U\cup S, & \text{if $0 \in U$}
\end{cases}$$
for some $U \subseteq \mathbb{R}$ which is open with in the standard topology on $\mathbb{R}$.
Now we have $\overline{\{0\}} = S \cup \{0\}$, since every neighbourhood of an element of $S$ contains $0$:
Take an arbitrary $x \in S$. For every neighbourhood $U$ of $x$ we have $\{0\}\cap(U\setminus\{x\})=\{0\}\ne\emptyset$.
