Compactness of [0, 1] in a specific topology

[radou]

Homework Statement

Let R have the topology consisting of all the sets A such that R\A is either countable or all of R. Is [0, 1] a compact subspace in this topology?

The Attempt at a Solution

If U covers R, if it consists of sets of type A such that R\A is finite, then [0, 1] is compact, since, if we take any set A from U, and if A doesn't cover [0, 1], then there's a finite number of points missing to cover [0, 1], let's say x1, ..., xn, and for each xi we can take a set Ai from U containing xi until we cover [0, 1], so we have a finite subcover (consisting of n+1 sets of U).

The interesting case is where R\A is countably infinite.

Let U be a cover for [0, 1] consisting of the following sets:

U1 = R\{1/n : n is a positive integer}
U2 = R\{1/(n+1) : n is a positive integer}
U3 = R\{1/(n+2) : n is a positive integer}, etc.

The family U covers [0, 1], right? But it doesn't have a finite subcover. Does this work?

Edit: by the way, Munkres says:

Definition. If Y is a subspace of X, a collection A of subsets of X is said to cover Y if the union of its elements contains Y.

Lemma 26.1. Let Y be a subspace of X, Then Y is compact if and only if every covering of Y by sets open in X contains a finite subcollection covering Y.

So I considered sets open in R, in the given topology. I hope there's no "subspace issue" I ignored here unwillingly.

 

[mighty2000]

Your solution looks correct to me. You have correctly identified that the interesting case is when R\A is countably infinite, and you have provided a cover for [0, 1] that does not have a finite subcover. This shows that [0, 1] is not compact in the given topology.

As for the subspace issue, I don't think there is one here. You have defined the topology on R, and [0, 1] is a subset of R, so [0, 1] is also a subspace of R. The definition and lemma you have provided are for a subspace Y of a larger space X, so there is no issue there.

Overall, your solution is well-written and clearly explains your reasoning. Good job!