I came across the comments here: Prove that $\inf\left\{t>\sigma:X_t=\varepsilon\right\}$ is a stopping time
and I'm curious about finding $\inf([\sqrt{2},2] \cap \mathbb{Q})$
Please, tell me if I'm wrong (I want to clarify my understanding of supremum and infimum): I think the problem here is in the definitions:
Let be the set: $$U = \{x \in \mathbb{R}; x = inf([\sqrt{2},2] \cap \mathbb{Q}])\}$$ In this case, the comment given by Did is right, we have $U = \{\sqrt{2}\}$
But if I define: $$U = \{x \in \mathbb{Q}; x = inf([\sqrt{2},2] \cap \mathbb{Q}])\}$$ Then 0xbadf00d is right: $U = \{\}$
For Every, $x\in [\sqrt{2},2]\cap{\Bbb Q}$ we have, $x\ge \sqrt2$
and for a given $\epsilon>0$ since, $\Bbb Q$ is dense in $\Bbb R$ there exists
$x_\epsilon\in [\sqrt{2},\min(2,\sqrt2+\epsilon)]\cap{\Bbb Q}$ that is, $x_\epsilon\in [\sqrt{2},2]\cap{\Bbb Q}$ and $$\sqrt2 \le x_\epsilon \le \sqrt2+\epsilon$$ this prove by definition that :$$\inf\{[\sqrt{2},2]\cap{\Bbb Q}\}=\sqrt{2}.$$