How would you prove that $\mathbb{Q} - B$, where $B$ is a finite subset of $\mathbb{Q}$, is dense in $\mathbb{R}$?

Question

I had a problem in my book asking to prove if various sets were dense in $\mathbb{R}$. This set I was working with ended up being equivalent to $\mathbb{Q} - \{0\}$ , and I realized $\mathbb{Q}$ already had an infinite number of holes in it with the irrational numbers and was still dense in $\mathbb{Q}$, thus missing the single point 0 would not change that.

I then realized as long as we weren't missing an interval $a$ to $b$, with $a<b$ and $a$ and $b$ being elements of $\mathbb{Q}$, even if we had an infinite subset removed from $\mathbb{Q}$ it should still be dense in $\mathbb{R}$. This would especially be true if the subset was finite.

I was wondering how we'd go about proving this idea, or even if I'm correct in thinking this.

Answer 1

In fact, you can take $B$ to be any set with no accumulation points.

Hint: Let $a<b$ be real numbers. First prove that there are infinitely many rational numbers in the interval $(a,b)$. Those cannot be all contained in $B$, otherwise $B$ would have an accumulation point in $(a,b)$.

EDIT: In fact, the same strategy also works under the weaker assumption that $B$ is nowhere dense, i.e. it closure has empty interior.

Answer 2

Prove that $\mathbb Q \setminus B$ is dense in $\mathbb R$ if $B$ is finite

Take $x\in \mathbb R$ and $\epsilon >0$. We have to prove that it exists $y \in (x-\epsilon, x+\epsilon) \cap \mathbb Q \setminus B$ with $y \neq x$.

Note $D = d(x, B\setminus\{x\})$. As $B$ is finite, we have $D>0$. Now as $\mathbb Q$ is dense in $\mathbb R$, it exists $y \in ( x-\epsilon^\prime,x+\epsilon^\prime) \cap \mathbb Q$ where $\epsilon^\prime=\inf(\epsilon,D)$ and $y \neq x$. Based on those conditions, we have $y \notin B$ which concludes the proof.

Answer 3

You have to show any open interval $I=(\alpha,\beta)$ contains points of $\mathbf Q\smallsetminus B$.

Either $B\cap I=\varnothing$, and it contains points of $\mathbf Q$ which cannot belong to $B$.

Or $B\cap I\neq\varnothing$. Let $b_1<\dots<b_k$ be its elements. The open intervals $(alpha, b_1)$ and $(b_1,\beta)$ are subsets of $I$, and contain elements of $\mathbf Q$, necessarily not in $B$.

Answer 4

Consider $x\in \mathbb{R}$ and a sequence of rationals $q_1,q_2.\cdots \in \mathbb{Q}$ such that $|q_n-x|\to 0$.

Suppose $\mathbb{Q}-B$ misses no interval as you state. Then, for every $n$ there must exist $p_n\in (\mathbb{Q}-B) \cap [q_n,q_{n+1}]$ (or $[q_{n+1},q_{n}]$ depending on which is greater). Now, $|p_n-x|<\max(|q_{n+1}-x|,|q_n-x|)\to 0$ so $p_n\to x$ as desired.

Answer 5

Given any real number $x$, there exists a sequence of rationals, $(q_n)$ such that $q_n\to x$. We can safely remove any finite number of points from a sequence without affecting the limit. The result follows.