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I'm looking for some examples of non-spatial frames. (a frame is non-spatial iff not isomorphic to any frames have forms of topologies for some sets)

A simpler example is better for me.

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In this excellent answer you have a characterization, and a sketch of a proof, of those lattices which are the lattices of open sets of some topology.

It tells that those are the complete lattices with plenty of prime elements.
(Here an element $p \neq 1$ is said to be prime if $p \leq a \wedge b$ always implies that $p \leq a$ or $p \leq b$.)
Having plenty of prime elements means that if $a \nleq b$ then $b \leq p$ and $a \nleq p$, for some prime element $p$.
The join-infinite distributive law (JID), $$x \wedge \bigvee_{i \in I}y_i = \bigvee_{i \in I}(x \wedge y_i)$$ is a consequence of these.

Now to get the example you ask for, we need a complete lattice satisfying (JID) but not having enough prime elements.
An example without prime elements at all will suffice.
Notice that, in a distributive lattice, a prime element will be meet-irreducible, that is, if $a \wedge b = p$, then $a = p$ or $b = p$.
In particular, in a Boolean algebra, a prime element is a co-atom (an element that is covered by the top element).
A Boolean algebra has co-atoms iff it has atoms (which are they're complements), so a complete atomless Boolean algebra will be an example of a non-spatial frame (notice that a complete Heyting algebra is a frame, and a Boolean algebra is a particular case of a Heyting algebra).


In case you don't know about any complete atomless Boolean algebra, in this answer there is a description of one such algebra, the one of the regular open sets of $\mathbb R$ (relative to the usual topology).

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