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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)

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    $\begingroup$ Your reasoning for $A$ being compact seems incorrect. $A = \{1, \ldots, n\}$ is compact because given an open cover $\{ B_\alpha \}$ of $A$, we can find $B_{\alpha_k} \ni k$ for each $k \in A$ so that $\{ B_{\alpha_k} \}_{k=1}^n$ covers $A$. $\mathbb{N}$ is not compact in your topology because the infinite cover $\{ \{1, \ldots, n\} \}_{n=1}^\infty$ of $\mathbb{N}$ cannot be reduce to a finite subcover. $\endgroup$ – parsiad Apr 8 at 1:33
  • $\begingroup$ @parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $\Bbb N$ is not compact as given an open cover $\{B_\alpha\}$ of $\Bbb N$ then for each $n\in \Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ? $\endgroup$ – excalibirr Apr 8 at 1:38
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    $\begingroup$ I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $\mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition. $\endgroup$ – parsiad Apr 8 at 1:39
  • $\begingroup$ @parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ? $\endgroup$ – excalibirr Apr 8 at 1:48
  • $\begingroup$ Yes, I think that part of your argument is correct. I posted a detailed answer anyways. $\endgroup$ – parsiad Apr 8 at 2:06
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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.

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$A=\{1\}$ 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.

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  • $\begingroup$ To the proposer: No infinite $B\subset \Bbb N$ is compact in this topology because $\{\{1,b\}: b\in B\}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $\Bbb N,$ a subset is compact iff it is finite. So if $1\in C\subset \Bbb N$ and $C$ is finite then $C$ is compact but $\overline C=\Bbb N$ is not compact. $\endgroup$ – DanielWainfleet Apr 8 at 6:37

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