Actually, let us define an operation, called bracket, that takes a relation \$$R\$$, and two elements in a set, and gives \$$\mathrm{true}\$$ when \$$R\$$ holds for the elements \$$x\$$ and \$$x'\$$ and \$$\mathrm{false}\$$ otherwise,

\$[x R x'] =\begin{cases}\mathrm{true} & \mathrm{if} \ x R x' \\\ \mathrm{false} & \mathrm{otherwise} \end{cases} \$

then,

\$\cap_X(x,x') := [x \leq x']. \$

Interesting fact about brackets is that when we identify \$$\mathbf{Bool}\$$ with \$$(\varnothing \subseteq \mathbf{1})\$$, with \$$\mathbf{false} \cong \varnothing\$$ and \$$\mathbf{true} \cong \mathbf{1}\$$ then taking the cartesion product with them allows us to have an object appear only when our relation holds (up to isomorphism),

\$[x R x'] \times A \\\\ \implies A \text{ when } [x R x'] = \mathrm{true} \\\\ \implies \varnothing\text{ when } [x R x'] = \mathrm{false} \$

since the cartesion product of any \$$A\$$ with \$$\varnothing\$$ is isomorphic to \$$\varnothing\$$, and the cartesion product of any \$$A\$$ with \$$\mathbf{1}\$$ is isomorphic to \$$A\$$.

\$(\cup\_{x,x} \times id_{x'}) \circ (id_x \times \cap_{x,x')} ) \\\\ = ([x \geq x]\times id_{x'}) \circ (id_x \times [x \leq x'] ) \\\\ = ([x = x]\times id_x') \circ (id_x \times [x \leq x'] ) \$
is equivalent to an upper-set, if we are allowed to vary \$$x'\$$. Which is sort of what I wanted to show above.