Failure of the law of excluded middle in geometric logic

Question

I am reading Steven Vickers' Topology via Logic, but I am struggling with exercise 3.5, which reads

For negation $\neg x$ in a frame, show that $x \land \neg x = \text{false}$ but find an example in $\Omega \mathbb{R}$ for which $x \lor \neg x \neq \text{true}$.

Here, $\Omega \mathbb{R}$ is a topology on the reals (if I understand correctly), meet $\land$ is defined as $\bigwedge X = \text{int}(\bigcap X)$, join $\lor$ is simply union, and $\neg x$ is defined as $$\neg x = x \rightarrow \text{false} = \bigvee \{c \mid c \land x \leq \text{false}\}$$

I think I am able to show the first part in the following way: $$x \land \neg x = x \land \bigvee\{c \mid c \land x \leq \text{false}\} = \bigvee(\{c \mid c \land x \leq \text{false}\} \land x) = \bigvee \text{false} = \text{false}.$$

However, I don't know how to show that $x \lor \neg x \neq \text{true}$. I think I need to somehow use the fact that the meet in the definition of negation is in fact the interior in order to provide a counterexample, but I don't see how to do it. Can someone help me?

Answer 1

Observe that for an open $O\subseteq {\mathbb R}$ your description of the negation concretely means that $\neg O = \text{int}({\mathbb R}\setminus O)$, the interior of the complement of $O$. With this description, you'll see that $O\vee\neg O\neq {\mathbb R}$ in general - in fact, for any $O\notin\{\emptyset, {\mathbb R}\}$ by the connectedness of ${\mathbb R}$. Let me know if you need details.