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)

• 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. – parsiad Apr 8 at 1:33
• @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 ? – excalibirr Apr 8 at 1:38
• 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. – parsiad Apr 8 at 1:39
• @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 ? – excalibirr Apr 8 at 1:48
• Yes, I think that part of your argument is correct. I posted a detailed answer anyways. – parsiad Apr 8 at 2:06

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.

$$A=\{1\}$$ 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.

• 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. – DanielWainfleet Apr 8 at 6:37