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

Question

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?

Thanks in advance!

Answer 1

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.

Answer 2

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.

Answer 3

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 $[0,1]$.

Answer 4

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.