Find an open cover that doesn't contain a finite subcover

Question

I need to solve the following question:

Find an open cover for $\mathbb{Q} \cap [0,1] \subset (\mathbb{R}, T_{st})$ that doesn't contain a finite subcover, where $T_{st}$ is the standard topology.

Q: How do I solve this?

I've tried to solve this question but I didn't succeed. Help would really be appreciated, thanks in advance!

Answer 1

First, some motivation. It is a fact that a compact subset of a Hausdorff space must be closed. Try to prove this. If you can, then try to reengineer the proof to see why the non-closed set $\mathbb{Q} \cap [0,1]$ cannot be compact. If you need more guidance, see below.

Choose an irrational number $\alpha \in (0,1)$. For each $x \in \mathbb{Q} \cap [0,1]$, we can choose disjoint open sets $U_x \ni x$ and $V_x \ni \alpha$. The open cover $\{U_x\}_{x \in \mathbb{Q} \cap [0,1]}$ does not have a finite subcover. (Why?)

Answer 2

Pick an irrational $\alpha$ in the unit interval. For each rational $r$ find an open interval $U_r$ that contains $r$ but avoids $\alpha$. Then the collection $\{U_r\}$ form an open cover of the rationals. But any finite subcollection will fail to cover the rationals, since the union of a finite number of intervals $U_r$ will stay a positive distance away from $\alpha$; in that nonzero gap will be at least one rational.

Answer 3

Thought process: $[0,1] $ is compact in $\mathbb R $ because it is closed and bounded. So why isn't $[0,1] $ in $\mathbb Q $?

The set bounded. And it's closed in $\mathbb Q $ but it is not closed as a subset of $\mathbb R $.

So what difference does not being closed or being in the space $\mathbb Q $ have to do with anything.

Well $[0,1]\cap \mathbb Q $ does not contain all its limit points. And $\mathbb Q $ as a space isn't complete; that is $\mathbb Q $ has cauchy sequences that don't converge to rational points. I.e those which, when viewed in the real space converge to an irrational point.

So $[0,1] \cap \mathbb Q$ not being compact must have something to do with irrational limit points.

So pick one. Call it $x\in [0,1] $. There are infinite sequence that converge to $x $ with no finite subsequence converging to $x $. so we can have one sets that cover everything arbitrarily close but never containing $x $, i.e. we can make an open cover of all but $x$.

So example: $\{\{y: y> x+\epsilon or y <x-\epsilon;\epsilon >0\}\} $ would be an open cover with no finite subcover.

Answer 4

Let $r$ be an irrational member of $(0,1).$

Let $(a_n)_{n\in \mathbb N}$ be a strictly increasing sequence of irrational members of $(0,r)$ with $\lim_{n\to \infty}a_n=r$. (E.g. $a_n=r(1-2^{-n}).\;$).

Let $(b_n)_{n\in \mathbb N}$ be a strictly decreasing sequence of irrational members of $(r,1)$ with $\lim_{n\to \infty}b_n=r.$

Let $C=\{[0,a_1)\cup (b_1,1]\}\cup \{(a_n,a_{n+1}): n\in \mathbb N\}\cup\{(b_{n+1},b_n):n\in\mathbb N\}.$

Let $D=\{c\cap \mathbb Q:c\in C\}$.

Then $D$ is an infinite open cover of the space $[0,1]\cap \mathbb Q$ and no proper subset of $D$ is a cover of $[0,1]\cap\mathbb Q.$