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)

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.