Arsthbt

Could someone tell me if my line of reasoning is correct here:

Say we have the topological space $(mathbbN, T)$ 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 $overlineA = Bbb N$

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)

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 $mathscrB equiv B_alpha$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscrB$ 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 $mathscrC equiv C_n_n=1^infty$ of $mathbbN$.
Let $mathscrC^prime$ be a finite subcollection of $mathscrC$.
Note that the set $bigcup_C in mathscrC^prime C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbbN$ (because none of its members contain $N+1$).
This shows that $mathbbN$ 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, $overlineA = mathbbN$.

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.

$A=1$ is compact as any cover of it has a one-element subcover.

$overlineA = mathbb N$ 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.