Question on compact metric space.

Question

Is the set of rationals in $[0,1]$ compact?

I seems like every open covering should have a finite sub-covering, yet I have read that compact metric spaces are complete...

Answer 1

No, it is not compact: consider the open cover

$$\left\{\Bbb Q\cap\left[0,\frac{\sqrt2}2-\frac1n\right):n\ge4\right\}\cup\left\{\Bbb Q\cap\left(\frac{\sqrt2}2+\frac1n\right]:n\ge4\right\}\;.$$

Answer 2

Hints:

Define

$$a_n:=\sum_{k=1}^n\frac1{2k^2}\implies\;\text{the sequence}\;\;\{a_n\}_{n\in\Bbb N}\subset K:=\Bbb Q\cap[0,1]\;\;\text{converges, yet}$$

$$\lim_{n\to\infty}a_n=\frac{\pi^2}{12}\notin K\;,\;\;\text{and thus}\ldots$$

Answer 3

It's worth noting that if it were compact, then it would be compact in $\Bbb R$, since the inclusion map $\Bbb Q\hookrightarrow\Bbb R$ is continuous, and the continuous image of a compact set is compact. But its image is $\Bbb Q\cap[0,1]$--where $[0,1]$ denotes the real interval, here. This is not closed in $\Bbb R,$ so not compact in $\Bbb R$. Thus, by contrapositive, the rational interval $[0,1]$ is not compact in $\Bbb Q$.

Answer 4

Use the fact that in a compact metric space, every sequence has a convergent subsequence. But , if we take the sequence : $a_n:=1,1.4,1.414,... $ , where $a_n$ has the first n terms of the decimal expansion of $2^{1/2}$ . This sequence will not converge in $\mathbb Q$, because $2^{1/2}$ is irrational. The same can be said for any irrational number, so that you can construct uncountably-many sequences that do not converge.

Answer 5

Another possible answer:

If $\mathbb{Q}\cap[0,1]$ were compact, the inclusion into $\mathbb{R}$ would obtain its maximum. Does it?