# Proving $[0,1]\cap\mathbb{Q}$ is not compact in $\mathbb{Q}$ [duplicate]

Is $$[0,1]\cap\mathbb{Q}$$ a compact subset of $$\mathbb{Q}$$.

No it is not. Since $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ for any open set in [0,1] there is a rational that belongs to that set. Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite, since it be composed of infinite number of rationals.

Question:

Is this proof right?

## marked as duplicate by Brahadeesh, Stahl, Watson, Alexander Gruber♦Nov 30 '18 at 3:30

• No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals. – freakish Nov 28 '18 at 18:24
• But $\left[ 0, 1 \right]$ is compact in $\mathbb{R}$ with the usual topology. Now, the set $\left[ 0, 1 \right] \cap \mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $\mathbb{R}$. Also, it will be compact in the relativised topology on $\mathbb{Q}$. – Aniruddha Deshmukh Nov 28 '18 at 18:25
• @AniruddhaDeshmukh $[0,1] \cap \mathbb Q$ is not closed in $\mathbb R$, therefore not compact in $\mathbb R$. If it were compact in $\mathbb Q$, it would be compact in $\mathbb R$. – Robert Israel Nov 28 '18 at 18:40
• Yes. I was wrong there! – Aniruddha Deshmukh Nov 28 '18 at 18:42
• "Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid. – fleablood Nov 28 '18 at 18:58

We know not-closed sets are not compact. Why? Because if the set is not closed there is a limit point not in it. And we can have infinite number of open sets that cover everything "up to" the limit point and there can't be a finite subcover as there will be a measurable quantifiable "gap" between the "largest" interval and the limit point that will not be covered.

So let's take some limit point in $$[0,1]\cap \mathbb Q$$ that is not in $$[0,1]\cap Q$$; in other words any irrational number between $$0$$ and $$1$$. Let $$x; 0 < x < 1$$ and $$x$$ is irrational.

We can take the open sets $$U_i= (-1, x-\frac 1i)\cup (x+\frac 1i,2)\cap \mathbb Q$$. (As $$(-1, x-\frac 1i)\cup (x+\frac 1i,2)$$ is open in $$\mathbb R$$ we know $$(-1, x-\frac 1i)\cup (x+\frac 1i,2)\cap \mathbb Q$$ is open in $$\mathbb Q$$. And $$\cup U_i$$ covers $$[0,1]\cap \mathbb Q = ([0,x)\cup(x,1])\cap \mathbb Q \subset( (-1,x)\cup(x, 2))\cap \mathbb Q$$.

And $$\{U_i\}$$ has no finite subcover.

After that brainstorming we can put it in simpler terms:

For any irrational $$x: 0 < x < 1$$ then $$[0,x) \cup (x, 1]$$ is not compact in $$\mathbb R$$. An open cover that has no subcover would be $$\{V_i|V_i = (-1,x-\frac 1i)\cup (x+\frac 1i, 2); i \in \mathbb N\}$$.

If we restrict that to $$\mathbb Q$$ then $$\{U_i| U_i = V_i\cap \mathbb Q\}$$ is an open (in $$\mathbb Q$$) cover of $$([0,x) \cup (x, 1])\cap \mathbb Q$$. But $$([0,x) \cup (x, 1])\cap \mathbb Q= [0,1]\cap \mathbb Q$$ (because $$x\not \in \mathbb Q$$).

So $$[0,1]\cap \mathbb Q$$ is not compact.

• I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that $\{U_i\}$ has no finite subcover? Thanks in advance! – Pedro Gomes Nov 29 '18 at 12:24
• Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-\frac 1n, x + \frac 1n)$ is not covered. This is a very standard tool in a bag of tricks mathematicians develop. – fleablood Nov 29 '18 at 17:18

A useful property of compactness is that it is intrinsical. In our case $$\mathbb{Q}\cap [0,1]$$ being compact in $$\mathbb{Q}$$ is equivalent to it being compact in $$\mathbb{R}$$. Now the compact subsets of $$\mathbb{R}$$ are the closed bounded sets, and clearly our set is not closed, since its closure is $$[0,1]$$.

If you want to show that $$[0,1]\cap\mathbb{Q}$$ does have an infinite covering which cannot be refined to a finite one consider this:

Let $$a_n=\sqrt{2}/2-1/n$$. Note that each $$a_n$$ is irrational. Now consider $$I_n=(a_n,a_{n+1})$$ together with $$[0,a_1)$$ and $$(\sqrt{2}/2, 1]$$. All those sets together form an open covering of $$[0,1]\cap\mathbb{Q}$$ but there is no finite refinement. In fact every rational belongs to exactly one element from that set meaning there is no refinement (even infinite) at all.

Consider, for $$\varepsilon>0$$, $$A_\varepsilon=\bigl((-\varepsilon,\sqrt{2}/2-\varepsilon)\cup(\sqrt{2}/2+\varepsilon,1+\varepsilon)\bigr)\cap\mathbb{Q}$$ for $$n>0$$. Then $$\bigcup_{\varepsilon>0} A_\varepsilon\supseteq[0,1]\cap\mathbb{Q}$$ Is there a finite subcover?

More conceptually, a compact subset of a Hausdorff space is closed.