We know that $\mathbb{R}$ with the usual topology is not compact.
Do we know the compactness of $\mathbb{R}$ equipped with either co-finite or
co-countable topology?
I am stuck on constructing a proof in both cases and I need help.
Proof attempt:
(Educated Guess: since no one has ever made a big deal about these
spaces being compact, I guess they are not compact)Let $\mathbb{R}_{co-f}$ be the reals with the co-finite
topology. Suppose $\mathcal{U}$ is an open cover of
$\mathbb{R}_{co-f}$. Suppose $\mathbb{R}_{co-f}$ is compact. We wish to
derive a contradiction.Let $U = \{U_i |i \in F, F \text{ is a finite subset of }
\mathbb{R}_{co-f}\}$ be the finite subcover. Then since $\mathbb{R}_{co-f}$ is compact, $\bigcup_{i\in F}
U_i = \mathbb{R}_{co-f}$(Hmmmm….This proof is going no where)
Take an arbitrary open set $V$, then $\mathbb{R}_{co-f} \backslash V$ is finite. Then $\mathbb{R}_{co-f} \backslash V$ must be covered by some open sets in the subcover..Let $\{U_i| i \in I \subseteq F\}$ be the subcover that covers
$\mathbb{R}_{co-f} \backslash V$, whch means $\{\mathbb{R}_{co-f}
\backslash U_i| i \in I \subseteq F\}$ covers $V$. However, each
$\mathbb{R}_{co-f} \backslash U_i$ is a finite set. A finite union of
finite set is finite. But since $V$ is co-finite in $\mathbb{R}$,
which is uncountable, therefore $V$ cannot be covered by finite sets.Therefore $\mathbb{R}_{co-f}$ is not compact as claimed.
Next proof:
Claim: $\mathbb{R}_{co-c}$ is not compact.
Proof: This time take $V$ as above. Since $\mathbb{R}_{co-f}
\backslash V$ is countable, it cannot be covered by finite open sets.
Since $V$ is not finite, it cannot be covered by finite open sets
either.
Can someone verify if these are correct?
Best Answer
Any set $A$ endowed with the co-finite topology is compact.
Let $\mathcal F$ be an open cover for $A$, select $F\in \mathcal F$. Let $\{a_1,a_2\dots a_n\}=\mathbb R \setminus F$ and for each $i\in\{1,2\dots n\}$ let $F_{i}$ be an element of $\mathcal F$ that contains $a_i$
Then $\{F,F_1,F_2\dots F_n\}$ is a finite subcover for $A$.
$\mathbb R$ with the co-countable topology is not compact.
Consider the sets $F_n=\mathbb R\setminus \{n,n+1\dots\}$
Clearly $\mathcal F=\{F_1,F_2\dots \}$ is an open cover for $\mathbb R$
On the other hand it has no finite subcover, because such a subcover would be of the form $\{F_{a_1},F_{a_2}\dots F_{a_n}\}$ with $a_1<a_2<\dots< a_n$. Notice $a_n$ is in none of those sets.