A confusion over the lebesgue outer measure of $(0,1) \cap \mathbb{Q}^c$

Question

I want to compute the lebesgue outer measure of $(0,1) \cap \mathbb{Q}^c$ where $\mathbb{Q}$ is the set of rational numbers.

$(0,1) = \{(0,1) \cap \mathbb{Q}\} \cup \{(0,1) \cap \mathbb{Q}^c\}$

By sub-additivity of lebesgue outer measure, we have $m^*\{(0,1)\} \leq m^*\{(0,1) \cap \mathbb{Q}\} + m^*\{(0,1) \cap \mathbb{Q}^c\}$

$\Rightarrow m^*\{(0,1) \cap \mathbb{Q}^c\} \geq m^*\{(0,1)\}(=1)- m^*\{(0,1) \cap \mathbb{Q}\}(=0)$

$\Rightarrow m^*\{(0,1) \cap \mathbb{Q}^c\} \geq 1$

Again by monotonicity of lebesgue outer measure, $m^*\{(0,1) \cap \mathbb{Q}^c\} \leq m^*\{(0,1)\}(=1)$

$\Rightarrow m^*\{(0,1) \cap \mathbb{Q}^c\} \leq 1$

Thus we have $m^*\{(0,1) \cap \mathbb{Q}^c\}=1$, as expected.

$\textbf{My problem:}$ I encountered a theorem that states: $m^*(U)=m_{*,(J)}(U)$ for all open sets $U$. Now $(0,1) \cap \mathbb{Q}^c$ seems to be an open set. Since $\mathbb{Q}$ is countable, so is $(0,1) \cap \mathbb{Q}$. Then we can write $(0,1) \cap \mathbb{Q}$ as $\{q_1,q_2, \ldots \}$. Thus we can write $(0,1) \cap \mathbb{Q}^c$ as $(0,q_1) \cup (q_1,q_2) \cup \cdots $, i.e. as countable union of open sets. Hence $(0,1) \cap \mathbb{Q}^c$ is open. Now $m_{*,(J)}\{(0,1) \cap \mathbb{Q}^c\}=0$. Then $m^*\{(0,1) \cap \mathbb{Q}^c\}=0$ (!!)

I maybe (must be!) completely off the track and I apologize in advance. Any clarification would be much appreciated!

Answer 1

$(0,1)\cap\mathbb Q^c$ is not open. Your attempt to write it as a union of open intervals (where your use of $\cap$ instead of $\cup$ is presumably just a typo) incorrectly presumes that $q_1<q_2<\dots$.

Answer 2

Since $\mathbb{Q}$ is dense in the real numbers, any open interval $(a,b)$ contains a rational. It is therefore impossible to obtain an open set consisting entirely of irrational numbers.