Confusion over the concept of "compactness"

Question

I have to prove some stuff that involves the concept of collection, in particular those relating to compact sets.

But then I have got this trouble. For example, consider the set of all rationals. If we take any rational $r$, certainly $\mathbb{R}$ has an open interval that contains $r$, so $\mathbb{R}$ covers $\mathbb{Q}$. But then there is one open set (the real numbers) that covers $\mathbb{Q}$ so $\mathbb{Q}$ is compact. I know this is not true, but I don't understand why this argument is incorrect.

Can somebody give some advice please? I tried hard to understand the problem but I can't see anything.

Answer 1

The problem here is that you misunderstand compactness.

A set $S$ is compact if and only if, for ANY covering of $S$ with open sets, there exists a finite subcovering which also covers $S$

Yes, indeed, $\mathbb R$ does cover $\mathbb Q$, meaning that $\{\mathbb R\}$ (i.e. the SET that contains one element, the set of real numbers) is a covering of $\mathbb Q$, and it is finite, therefore it is a finite covering of $\mathbb Q$.

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However, that does not mean that $\mathbb Q$ is compact. For example, the set $$C=\{(z-1, z+1)| z\in\mathbb Z\}$$

is also a covering of $\mathbb Q$, because $$\mathbb Q\subseteq \bigcup_{A\in C} A$$

However, it is clear to see that no finite subcovering of $S$ can ever hope to cover $\mathbb Q$, since any finite union of sets from $C$ will have an upper bound, and $\mathbb Q$ is unbounded.

Answer 2

A set is compact of for every open cover you can extract a finite subcover. The cover $\{ \mathbb{R} \}$ of $\mathbb{Q}$ is already finite (since it consists of only one set), but the one you constructed (open intervals around rationals) cannot be finite unless the rationals are finite.

To check compactness in $\mathbb{R}$, recall also the Heine-Borel theorem: the compact subsets of $\mathbb{R}$ are the closed and bounded subsets.