Real line $\mathbb R^1$ with the countable complement topology is not compact [duplicate]

The real line with the countable complement topology is not compact.

I create an open cover of $$\mathbb R$$ by: Let $$\mathbb Q$$ denote the rationals. Let an open cover be defined be the set of $$\mathbb Q-\{q\}$$, where $$q \in \mathbb Q$$. Then this set is countable. I am unsure how to proceed from here. And is this a good example of an open cover that doesn't have a finite subcover? Which will lead to my conclusion of not compact?

marked as duplicate by Asaf Karagila♦ general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 3 '18 at 0:28

• Is "$Q\sim q$" the set of rationals other than $q$? If so, this doesn't work for three reasons: $(1)$ the union of all these sets is just $\mathbb{Q}$, but you want a cover of $\mathbb{R}$, $(2)$ each of these sets is countable and nonempty, so is not open in the countable complement topology on $\mathbb{R}$, and $(3)$ any two distinct elements of the family cover all of $\mathbb{Q}$. – Noah Schweber Dec 2 '18 at 20:33

The following approach is more work than necessary, but is a useful exercise in working with topological concepts:

• Every countable $$C\subset\mathbb{R}$$ is closed with respect to the countable complement topology.

• Every countable $$C\subset\mathbb{R}$$, with the subspace topology coming from the countable complement topology on $$\mathbb{R}$$, is discrete (= all sets open).

• If $$C\subset\mathbb{R}$$ is countable and infinite (some texts include infinitude in the definition of countability, but not all), then - by the previous bulletpoint - $$C$$ is not compact with respect to the subspace topology coming from the countable complement topology on $$\mathbb{R}$$.

• A general topological fact: If $$(X,\tau)$$ is a topological space, $$A\subset X$$ is closed with respect to $$\tau$$, and $$A$$ is not compact with respect to the subspace topology coming from $$\tau$$, then $$(X,\tau)$$ isn't compact either.

This is admittedly much harder than simply constructing a counterexample to compactness, which Jeffery has done. However, this argument has the advantage of giving you more general practice with multiple of the concepts involved: closed sets, subspace topologies, and open covers. It's definitely not the first way you should solve this problem, but you should go through the work above after you understand why $$(\mathbb{R},cocountable)$$ isn't compact.

Without even going so far as using rational numbers, what about this counter example: Define the open set $$X_i = \mathbb{R}-\mathbb{N}+\{i\}$$ with $$i\in \mathbb{N}$$. $$\cup_{i\in\mathbb{N}}X_i$$ is an open cover of $$\mathbb{R}$$ with no finite subcover.

Take any countably infinite subset (faithfully indexed) $$\{x_n: n=0,1,2,3,\ldots\}$$ of $$X$$ and define $$F_n =\{x_k: k \ge n\}$$, all of which are countable hence closed and obey the finite intersection property but whose intersection is empty. So no infinite set in the cocountable topology can even be countably compact.