Example of closed and bounded in $\mathbb Q$ set that is not compact

Question

Consider the metric space $Q$ of rational numbers with the Euclidean metric of $R$. Let $S$ consist of all rational numbers in the open interval ($a, b$), where $a$ and $b$ are irrational. Then $S$ is a closed and bounded subset of $Q$ which is not compact.

I am trying to prove this statement. It's easy to show that $S$ is closed and bounded but I'm having trouble showing that it is not compact. I think I need to find an example of an open covering of $S$ which doesn't have any finite subcover, but I can't think of such an example. Any help? Thanks.

Answer 1

HINT: Pick a sequence $a_n$ converging to $a$ from above, and consider $(a_n,b)$ as the open cover.

Another approach would be to prove that $(a,b)\cap\Bbb Q$ is not complete and that a compact metric space is always complete.

Answer 2

Consider $S=(a,b) \cap \mathbb Q$ as a subset of $\mathbb R$. Then $S$ is compact relatively to $\mathbb Q$ if and only if $S$ is compact relatively to $\mathbb R$. Clearly $S$ is not compact in $\mathbb R$, then $S$ is not compact in $\mathbb Q$.

Answer 3

This is not an answer but an addendum to Diego's answer, for later readers who need an explanation of "$S$ is compact relatively to $\mathbb Q$ if and only if $S$ is compact relatively to $\mathbb R$".

Let $(X,\tau_X)$ be a topological space (like $\Bbb R$) and $Y\subset X$ (like $\Bbb Q$), equipped with the induced topology $\tau_Y:=\{Y\cap U\mid U\in\tau_X\}.$ By definition, a subset $S\subset Y$ is compact relatively to $Y$ iff every open cover of $S$ in $Y$ has a finite subcover, i.e. iff for every $(U_i)_{i\in I}$ in $\tau_X$ such that $S\subset\bigcup_{i\in I}(Y\cap U_i),$ there is a finite $J\subset I$ such that $S\subset\bigcup_{i\in J}(Y\cap U_i).$ Since $S\subset Y,$ this can be simplified to: for every $(U_i)_{i\in I}$ in $\tau_X$ such that $S\subset\bigcup_{i\in I}U_i,$ there is a finite $J\subset I$ such that $S\subset\bigcup_{i\in J}U_i,$ which is precisely the definition of "$S$ is compact relatively to $X$".

Another point of vue is to realize that:

• compactness is an absolute topological notion: a subset $S$ of a topological space $(Z,\tau_Z)$ is compact relatively to $Z$ iff it is compact relatively to itself when equipped with the induced topology $\tau_S=\{S\cap U\mid U\in\tau_Z\}$;
• inducing topologies is "transitive": in the previous context $S\subset Y\subset X,$ the topology on $S$ induced from $\tau_Y$ which was itself induced from $\tau_X$ coincides with the topology on $S$ directly induced from $\tau_X.$