Prove: $\inf\big((x,y)\cap \mathbb{Q}\big)=x$ when $x,y\in \mathbb{R}$ and $x<y$

Question

I know that $\mathbb{Q}$ is dense in $\mathbb{R}$. However, I don't know what I am supposed to do. Any impulses?

Answer 1

The infimum is the largest lower bound for $(x,y) \cap \mathbb{Q}$. So you need to prove two things:

a) $x \le z$ for all $z \in (x,y) \cap \mathbb{Q}$, and

b) if $l \le z$ for all $z \in (x,y) \cap \mathbb{Q}$, then $x \ge l$.

I leave proving a) for you to do on your own.

For b), a suggested outline is proof by contrapositive. Suppose $x < l$. Then prove that there exists some $z \in (x,y) \cap \mathbb{Q}$ such that $l > z$. Denseness of $\mathbb{Q}$ in $\mathbb{R}$ might be helpful, as you already observed.

Answer 2

Let $S=(x,y)\cap \Bbb Q$ where $x<y$. $S$ is non-empty. Indeed, as you know that $\Bbb Q$ is dense in $\Bbb R$, you know that there exists $q\in\Bbb Q$ with $x<q<y$. Clearly, $x$ is a lower bound for $S$. As $S$ is non-empty and bounded from below $s:=\inf S$ is a real number. In fact, with the above, we have $x\le s\le q$.

Suppose $s>x$. Then again by density of $\Bbb Q$ in $\Bbb Q$, there exists $q'\in \Bbb Q$ with $x<q'<s<y$. It follows that $q'\in S$,contradicting the fact that $s$ is a lower bound of $S$. We conclude $s=x$.