Non-compactness of $\mathbb{R}$ with the cocountable topology

Is $(\mathbb{R},\tau_{co})$ compact where $\tau_{co}$ is the cocountable topology on $\mathbb{R}$?

I have the answer of my teacher but I'd like to see another one so I can understand better how people find the answer intuitively. He says:

$\mathbb{N}-\{n\}$ is infinite countable for all $n\in\mathbb{N}$. Then $V_n = \mathbb{R}-(\mathbb{N}-\{n\})=(\mathbb{R}-\mathbb{N})\cup\{n\}\in\tau_{co}$ for all $n\in\mathbb{N}$.

Then $U=\{ V_n : n\in\mathbb{N}\}$ is a open covering of $(\mathbb{R},\tau_{co})$.

$U$ does not contain a countable subcover since $V_n\cap\mathbb{N}=\{n\}$ for each $n\in\mathbb{N}$ and $\mathbb{N}$ is infinite.

How would you solve that problem?

• From Wikipedia: The cocountable topology on an uncountable set is hyperconnected, thus connected, locally connected and pseudocompact, but neither weakly countably compact nor countably metacompact, hence not compact. Commented Aug 21, 2014 at 2:59
• Cover $\mathbb R$ by open sets \{\mathbb R- \mathbb Z_{\geq n}:n\geq 0\} where $\mathbb Z_{\geq n}$ is the set $\{ m\in \mathbb Z : m\geq 0\}$. Commented Aug 21, 2014 at 3:03
• Proof by Wikipedia! :-) Commented Aug 21, 2014 at 3:07
• I'll note that $\mathbb{R}$ is compact in the co-finite topology, but only Lindelof in the co-countable topology. Why might this be? Commented Aug 21, 2014 at 3:14
• This is a fairly standard technique in exhibiting failure of compactness. The broad stroke techniques are:(a) find a countable or larger family of open sets covering everything but such that each of them is "indispensable" in the sense that there is a point it covers that no other covers, or (b) finding a family such that any finite subfamily blatantly fails to cover the whole thing, specifically an increasing sequence of sets $U_1 \subset U_2\subset ...$ such that no $U_n$ is the whole space, but the union is (balls of increasing integer radius in the plane for example).
– jxnh
Commented Aug 21, 2014 at 3:15

For $q \in \mathbb Q$, let $\mathcal O_q = (\mathbb R \setminus \mathbb Q) \cup \{q\}$ i.e. the set of all irrational numbers and $q$. Then, $\mathcal O_q$ is cocountable and $\mathbb R = \bigcup_{q\in \mathbb Q} \mathcal O_q$ but no finite subcover would cover all of the rational numbers.

• Ok, that's almost the same solution. Is it made by you or it is just like a standard method? I can't get the solution for simple topology problems and I don't know why. Commented Aug 21, 2014 at 3:05
• We are looking for cocountable sets and to find them we need countable sets first. Integers and rational numbers are kind of canonical choices as countable sets.
– user21965
Commented Aug 21, 2014 at 3:15

Another way is to make use of the characterisation of compactness by families of closed sets with the finite intersection property:

A topological space $X$ is compact iff whenever $\mathcal{A}$ is a family of closed subsets of $X$ with the finite intersection property (i.e., the intersection of any nonempty finite subfamily of $\mathcal{A}$ is nonempty), then $\bigcap \mathcal{A} \neq \varnothing$.

Note that the closed subsets of $( \mathbb R , \tau_{\text{co}} )$ are exactly the countable subsets of $\mathbb{R}$ (and $\mathbb R$ itself). It follows that for each $n \in \mathbb{N}$ the set $A_n = \{ k \in \mathbb N : k \geq n \}$ is a closed subset of $( \mathbb R , \tau_{\text{co}} )$. It is easy to see that $\{ A_n : n \in \mathbb N \}$ has the finite intersetion property (if $n_1 < n_2 < \ldots < n_\ell$ are natural numbers, then $A_{n_1} \cap A_{n_1} \cap \cdots \cap A_{n_\ell} = A_{n_\ell}$ is nonempty). However $\bigcap_{n \in \mathbb{N}} A_n = \varnothing$.

This can be translated into essentially your instructor's proof, but in co-finite and co-countable topologies I find that it is often easier to work with the closed sets as opposed to the open sets.