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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?

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    $\begingroup$ $\mathbf{R}$ with co-finite is compact. $\mathbf{R}$ with co-countable is not compact. Neither of your proofs are correct. $\endgroup$ – Willie Wong Aug 1 '16 at 3:25
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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.

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  • $\begingroup$ Thanks you just saved my final exam $\endgroup$ – Shamisen Expert Aug 1 '16 at 3:34
  • $\begingroup$ No problem, good luck with your test. $\endgroup$ – Jorge Fernández Hidalgo Aug 1 '16 at 3:36
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The cofinite topology is compact, while the cocountable is not.

For the cofinite, let $\mathscr U$ be a collection of opens that cover $\Bbb R$, and take a non-empty $A\in \mathscr U$. Then the complement of $A$ is finite, so for each point $x_i$ in the complement, pick a $B_i\in \mathscr U$ with $x_i\in B_i$. Then $\{A,B_1,\ldots,B_n\}$ is a finite covering of $\Bbb R$.

For cocountable, let $\mathscr V =\{A_1,A_2,\ldots\}$ with $$A_i=(\Bbb R\setminus \Bbb N)\cup \{i\}$$Then $\mathscr V$ covers the real line, but no proper subset of $\mathscr V$ does, and specifically no finite subset (a subset of $\mathscr V$ not containing $A_n$ does not cover $n$).

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An equivalent def'n of compact in terms of closed sets is: $X$ is compact iff any non-empty closed family $F$ in $X$ with the F.I.P. satisfies $\cap F\ne \emptyset.$

Translation: A closed family in $X$ is a set of closed subsets of $X.$ And "$F$ has the F.I.P. (Finite Intersection Property )" means that any non-empty finite $G\subset F$ satisfies $\cap G\ne \emptyset.$

$\quad$1.For the co-countable topology on $\mathbb R,$ let $F=\{f_n:n\in \mathbb N\}$ where $f_n=\{x\in \mathbb N : x\geq n\}.$ Then $\quad \; F$ is a non-empty closed family with the F.I.P. but $\cap F=\emptyset.$

  1. For the co-finite topology on $\mathbb R,$ let $F$ be a non-empty closed family with the F.I.P. Consider the family $F^*$ of non-empty finite subsets of $F.$ For each $G\in F^*$ we have $\cap G\ne \emptyset$ (by F.I.P.), and $\cap G$ is finite (because $\cap G \subset f\in F$ for any $f\in G).$

    Take any $G\in F^*$ such that the number of members of $\cap G$ is the least possible. Then $\cap F\supset \cap G. $ Because if not, then there exist $f\in F$ and $x\in \cap G$ with $x\not \in f.$ But then $G\cup \{f\}\in F^*$ and $\cap (G\cup \{f\})$ has fewer members than $\cap G, $ which is impossible.... And so $$\cap F\supset \cap G \ne \emptyset.$$ (We actually have $\cap F=\cap G$.)

Any statement about open sets has a dual statement about their complements, the closed sets. Many of the dual forms are useful in themselves.

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