I am attempting to prove $ℚ$ does not have the Heine-Borel property. I thought to prove that $ℚ$ is not compact, such that every open cover of $ℚ$ does not contain a finite subcover.
How would I go about this? Would it be the same as finding an open cover of $ℚ ∩ [0, 1]$ that cannot be reduced to a finite subcover?
I'm new to these concepts, so I do not fully understand everything. Any help will be greatly appreciated. Thanks!
On
What about the open cover
$$\left\{\;A_n:=\Bbb Q\,\cap\,(n,n+1)\;\;,\;\;B_n=\Bbb Q\,\cap\,\left(n-0.01,\,n+0.01\right)\;\right\}\;,\;\;n\in\Bbb Z\;\;?$$
Can you see why not only you can't have a finite subcover of the above that'll cover all of $\;\Bbb Q\;$ , but in fact you cannot even take one single open set $\;A_n\,,\,\,B_n\;$ without leaving out at least one rational number?
On
You can immediately see that $\mathbb Q$ is not compact, since it is not bounded. So for instance the cover $\{(-n,n)\}_{n\in\mathbb N}$ does not admit a finite subcover.
When you consider $\mathbb Q\cap[0,1]$, you can use that it is not closed (since a compact subset of $\mathbb R$ is closed). Or you can construct an open cover that does not admit an open subcover, like $$\Big\{\Big(-1,\tfrac1{\sqrt2}-\tfrac1n\Big)\cup\Big(\tfrac1{\sqrt2},2\Big)\Big\}_{n\in\mathbb N}.$$
That would be one way to do it. Specifically, you can let $\alpha$ be an irrational in $\left(\frac12,1\right)$, and for $n\in\Bbb Z^+$ let $\alpha_n=\alpha-\frac1{n+1}$. Let $U_0=[0,\alpha_1)\cap\Bbb Q$, and for $n\in\Bbb Z^+$ let $U_n=(\alpha_1,\alpha_{n+1})\cap\Bbb Q$. Finally, let $V=(\alpha,1]$, and let
$$\mathscr{U}=\{V\}\cup\{U_n:n\ge 0\}\,.$$
Now show that $\mathscr{U}$ is an open cover of $[0,1]\cap\Bbb Q$ that has no proper subcover (and hence no finite subcover), even theough $[0,1]\cap\Bbb Q$ is a closed, bounded subset of $\Bbb Q$.