Compactness of a topology on the natural numbers.

Question

Given that $\tau = \{ A \subset \mathbb{N} : \text{either } A = \emptyset \text{ or } 1 \in A \}$, is the topological space $(\mathbb{N}, \tau)$ compact?

I've shown that it fails to satisfy the Bolzano Weierstrass property and hence cannot be compact:

consider $U_\alpha = \{ \alpha, 1 \}$ for $\alpha \in \mathbb{N}>1$, and $U_1 = \{ 1 \}$.

This is a countably infinite set of elements of $(\mathbb{N}, \tau)$. Consider the neighbourhood $U_1$ of $1$, this has no other elements than $1$, and hence $1$ fails the criteria to be an interior point.

Because of this, the topology does not satisfy the B-W property. All topologies satisfying the B-W are compact, and hence this topology is not compact.

I don't feel confident I've fully understood the definitions i've used here. It could be a case of the problem actually being as simple as it seems, but I'm concerned I've made a mistake somewhere. Any advice would be appreciated.

Answer 1

I'll try to write a hinting answer because I think you suggested a right idea.

You suspect that the space is not compact. You should construct/choose an open cover which will have no finite subcover.

Take the collection you proposed, $\{ U_n \}_{n\in \mathbb{N}}$ with $U_n:=\{1,n\}$. Note that $\cup_{n\in \mathbb{N}}U_n=\mathbb{N}$.

Is it possible that there exists a finite subset ${i_1,...,i_m}\subseteq \mathbb{N}$ such that $\cup_{j=1}^m U_{i_j}=\mathbb{N}$? If not, you have found an open cover with no finite subcover.

Answer 2

The Bolzano-Weierstrass criterion for compactness is only equivalent for metric spaces as far as I know. A metric space is compact iff. it is sequentially compact, but this should fail for some nonsequential spaces.

$1$ does not fail the criterion to be an interior point of a neighbourhood of $1$; by definition, $1$ is always an interior point for a neighbourhood of itself. $\{1\}$ is open in this space. That's weird if you're used to 'ordinary' topologies like the one on $\Bbb R$, but it's completely legitimate for a point to be open. This also has nothing to do with Bolzano-Weierstrass or compactness.

It is nevertheless true that this space is not compact.