Is the subspace of rational numbers in the usual space of real numbers compact?
I'm not exactly sure what this is asking. Is this asking if I can generate a cover using a finite amount of sets from the rational numbers to cover all of the real numbers? Apparently, the answer is yes, but I'm not sure I could find a way to cover the irrational numbers.
On
Consider just the set $\Bbb Q$ of rational numbers, with the usual topology, which is that open sets are intervals of the form $$(a,b) = \{ x\in \Bbb Q \mid a < x < b \}$$ and unions of those intervals.
The question is:
In this topology, is the entire space $\Bbb Q$ compact?
That is, if someone gives you a family $\mathcal F$ of open intervals whose union is all of $\Bbb Q$, can you guarantee to find a finite subfamily of $\mathcal F$ whose union is also all of $\Bbb Q$?
A subspace $A$ of $\mathbb{R}$ is compact, means that for every family of open subsets of $\mathbb{R}$ that covers $A$, we can find a finite subfamily that also covers $A$.
In $\mathbb{R}$ we know that $A$ compact implies that $A$ is bounded: we can take the family $(-n,n)$ for $n \in \mathbb{N}$; these have the whole of the reals as their union, so every $A$ is covered by it. A finite subfamily is covering the same as one of them (for the largest $n$), and so $A \subset (-n,n)$ for some $n$.
This doesn't happen for the rationals...