1. ## Finding Open Cover

Let $A = \mathbb{Q} \cap [0,1] \subset \mathbb{R}$. Give an example of an open cover of $A$ which has no finite subcover.

I was thinking, since it specifies exactly that $A \subset \mathbb{R}$, then do our open sets (that are in the open covers) have to be open in $\mathbb{R}$, or can they be open in $\mathbb{Q}$?

For example, if $x$ is any rational number in $[0,1]$ then the set $\{ x \}$ is open in $\mathbb{Q}$, and the collection of all sets {x} such that x is a rational number in [0,1] is an open cover which has no finite subcover. But is it permissible for my open sets to be open in $\mathbb{Q}$? Or do they have to be open in $\mathbb{R}$?

2. No, they have to be open in R precisely because it said " $\subset \mathbb{R}$". A set or real numbers is compact if and only if it is both closed and bounded. A clearly is bounded so the crucial point is that it is not closed as a set of real numbers. Why, exactly,is it not closed? Use that to find your open cover.

3. Yup, that's exactly what I did (before I saw your post :P ), and I found my open cover. Thanks halls.