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?

  • $\begingroup$ 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. $\endgroup$ – parsiad Aug 21 '14 at 2:59
  • $\begingroup$ Cover $\mathbb R$ by open sets \{\mathbb R- \mathbb Z_{\geq n}:n\geq 0\} where $\mathbb Z_{\geq n}$ is the set $\{ m\in \mathbb Z : m\geq 0\}$. $\endgroup$ – Rachmaninoff Aug 21 '14 at 3:03
  • $\begingroup$ Proof by Wikipedia! :-) $\endgroup$ – user4894 Aug 21 '14 at 3:07
  • $\begingroup$ 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? $\endgroup$ – Chris K Aug 21 '14 at 3:14
  • 3
    $\begingroup$ 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). $\endgroup$ – JHance Aug 21 '14 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.


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.

  • $\begingroup$ 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. $\endgroup$ – Danowsky Aug 21 '14 at 3:05
  • $\begingroup$ 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. $\endgroup$ – user21965 Aug 21 '14 at 3:15

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