# Endoskeleton of a poset is isomorphic to skeleton of the poset proof?

Let $$(X, \leq)$$ be a poset (note that $$\leq \; \subseteq X \times X$$). Show that any endoskeleton of $$(X, \leq)$$ is isomorphic to the skeleton of $$(X, \leq)$$.

Some definitions:

Clique of $$(X, \leq)$$: $$(Y, \leq')$$ is a clique of $$(X, \leq)$$ iff $$Y \subseteq X$$ and $$\leq'$$ is a restriction of $$\leq$$ and a clique.

Maximal Clique of $$(X, \leq)$$: A clique of $$(X, \leq)$$ that is maximal.

Skeleton of $$(X, \leq)$$: Let $$(X_i, \leq_i)$$ be the family of all maximal cliques of $$(X, \leq)$$. Then the skeleton of $$(X, \leq)$$ is $$(\cup_i X_i, \cup_i \leq_i)$$.

Endoskeleton of $$(X, \leq)$$: Let $$(X_i, \leq_i)$$ be the family of all maximal cliques of $$(X, \leq)$$. Then the endoskeleton of $$(X, \leq)$$ is $$(X',\leq')$$ s.t. $$\forall i \; X' \cap X_i \neq \varnothing$$ and $$\leq'$$ is a restriction of $$\leq$$.

Monotone function: Given two preorders $$(X, \leq)$$ and $$(Y, \leq')$$, a function $$f: X \rightarrow Y$$ is monotone iff $$\forall x,x' \in X \; x \leq x' \Rightarrow f(x) \leq' f(x')$$.

Isomorphic pair: Given two preorders $$(X, \leq)$$ and $$(Y, \leq')$$, the monotonic functions $$f: X \rightarrow Y$$ and $$g: Y \rightarrow X$$ are isomorphic if $$g \circ f = id_X$$ and $$f \circ g = id_Y$$.

Isomorphism: Two preorders are isomorphic to each other if there exists an isomorphism pair between them.

My attempt: I am unable to find two monotonic function between those sets which are inverses of each other. Have I misunderstood some definition or do the functions elude me?

• Just an FYI, a clique is the codiscrete or chaotic relation i.e. $\forall x,y \; x R y$.
– Anon
Mar 31, 2021 at 16:54
• I assume your $\phi$ is supposed to be the empty set, $\varnothing$. You can write this in tex as \varnothing. Apr 2, 2021 at 2:25

A partial order is a preorder that has no nontrivial cliques. If you had two distinct elements, $$x$$ and $$y$$, that formed a clique, then you would have $$x\leq y$$ and $$y\leq x$$, which contradicts the antisymmetry of the poset. Hence, every element of $$X$$ forms a trivial maximal clique, consisting of only itself. Then the skeleton of a poset and all of its endoskeletons (of which there is only one), is isomorphic to the poset itself.