Showing the closure of a compact subset need not be compact Could someone tell me if my line of reasoning is correct here:
Say we have the topological space $(\mathbb{N}, T)$ comprising of the empty set together with all subsets of $\Bbb N$ that contain the element $1$.
I want to show that the closure of a compact set in this topology need not be compact. 
Let $A \equiv \{1, ...., n\}$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $\Bbb N$ which is not 1.
Therefore, its closure is $\overline{A} = \Bbb N$
If the above is true then I get confused because it seems that $\Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $\Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)
 A: 
Definition. Let $A \subset X$ where $(X, \tau)$ is a topological space.
  We call $A$ compact if for each collection $\{ B_\alpha \} \subset \tau$ such that $\cup_\alpha B_\alpha \supset A$, we can find a finite subcollection $\{B_{\alpha_1}, \ldots, B_{\alpha_n}\}$ such that $B_{\alpha_1 } \cup \cdots \cup B_{\alpha_n} \supset A$. 
Remark. We call the original collection an open cover and the subcollection a finite subcover.

Consider the topology in your original question.
Let $A \equiv \{1, \ldots, n\}$ and $\mathscr{B} \equiv \{B_\alpha\}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $\mathscr{B}$ is a cover of $A$, we can find $\alpha_k$ such that $k \in B_{\alpha_k}$.
Therefore, $\{B_{\alpha_k}\}_{k=1}^n$ covers $A$, and hence $A$ is compact.
Next, let $C_n \equiv \{1,\ldots,n\}$ and consider the cover $\mathscr{C} \equiv \{C_n\}_{n=1}^\infty$ of $\mathbb{N}$.
Let $\mathscr{C}^\prime$ be a finite subcollection of $\mathscr{C}$.
Note that the set $\bigcup_{C \in \mathscr{C}^\prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $\mathbb{N}$ (because none of its members contain $N+1$).
This shows that $\mathbb{N}$ is not compact.
Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $\overline{A} = \mathbb{N}$.
In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.
A: $A=\{1\}$ is compact as any cover of it has a one-element subcover.
$\overline{A} = \mathbb N$ which is not compact, as witnessed by the open cover $$\{\{1,2\},\{1,3\},\{1,4\},\ldots, \{1,n\}, \ldots\}$$ of $\mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.
