# Prove $\mathbb{Q} \cap [0, 1] \subseteq \mathbb{R}$ is not compact

Prove $\mathbb{Q} \cap [0, 1] \subseteq \mathbb{R}$ is not compact

To do this, all I need to do is construct on open cover $\mathcal{O}$ of $\mathbb{Q} \cap [0, 1]$ that doesn't contain a finite subcover.

I figured the easiest way to do this would be to construct $\mathcal{O}$ by taking irrational endpoints for the intervals, but I'm not quite sure how to construct it correctly. Any hints to this specific construction would be appreciated.

• – carmichael561 Apr 12 '17 at 1:53

alternatively construct a sequence with no convergent subsequence. Let $x_n$ be the first $n$ terms of the decimal expansion of $1/\sqrt{2}$ this does it.

Take $r = \frac{1}{\sqrt{2}} \in [0,1]$, which is irrational. Take an increasing sequence of rationals $q_n$ $q_n \in [0,1]$ such that $q_n \uparrow r$ (the truncated decimal expansions, e.g.).

Define $U_n = [0,1] \cap \mathbb{Q} \cap (\leftarrow, q_n)$, $V = [0,1] \cap \mathbb{Q} \cap (r, \rightarrow)$. These are by definition open in $\mathbb{Q} \cap [0,1]$ (interesections of open sets in $\mathbb{R}$ with $[0,1]\cap \mathbb{Q}$. As the $q_n$ are increasing, $U_n \subset U_{n+1}$ for all $n$.

$\{V\} \cup \{U_n: n \in \mathbb{N}\}$ is an open cover of $[0,1]\cap \mathbb{Q}$. If not, there would be a rational number $x \in [0,1] \cap \mathbb{Q}$ such that $x \le r$ (As $x \notin V$) and $x \ge q_n$ (for all $n$, as $x \notin U_n$, the latter forces $x \ge r$ and so $x=r$ a contradiction, as $r$ is irrational.

There is no finite subcover: suppose there were, say $\{U_{n_1}, \ldots, U_{n_k}, V\}$ is a finite subcover. Let $N$ be the maximal index occurring under $\{n_1,\ldots n_k\}$, then $U_{n_i} \subseteq U_N$, as the $U_n$ are increasing in $n$. If we pick any rational $q \in (q_N, r)$ (any non-empty open interval contains rational numbers), then $q \notin U_N$, so $q \notin U_{n_i}$ for $I= 1,\ldots k$ as well, and $q \notin V$, so the supposed subcover is not a cover of $[0,1]\cap \mathbb{Q}$ at all, a contradiction.

Take $r\in (0,1)$ \ $\mathbb Q.$

Let $(a_n)_{n\in \mathbb N}$ be a strictly increasing sequence of irrationals converging to $r,$ with $a_1>0.$ For example, $a_n=r(1-2^{-n}).$

Let $A_1=(-1,a_1)$ and $A_{n+1}=(a_n,a_{n+1}).$

Then $\{A_n:n\in \mathbb N\}\cup \{(r,2)\}$ is an open cover of $[0,1]\cap Q$ with no finite sub-cover.

It suffices to show that said set is not closed (because compact subsets of hausdorff-spaces are always closed). For this take a sequence of positive rationals converging to $\sqrt2/2$ from below. Then all the members of our sequence are in our set, but the limit is not, hence our set is not closed and therefore not compact.