# Dual cone operation is an antitone morphism of lattices

Currently I'm reading a classical text Convex Sets by F. Valentine. While reasing it I'm trying to generalize results as much as I can. The chapter on Dual cones made me thinkin of the fallowing theorem: Dual cone operation is an antitone morphism of lattices. Let me explain what I mean by this:

Let $$V$$ be a real vector space. Denote by $$\mathsf{CONV}(V)$$ set of all convex sets of $$V$$, and by $$\mathsf{CONE(V)}$$ set of all closed cones of $$V$$ with the origin at $$0$$. Note, that both $$\mathsf{CONV}(V)$$ and $$\mathsf{CONE(V)}$$ can be be endowed with the structure of non-distributive commutative lattice with meets operation $$(\wedge)$$ defined as intersection $$(\cap)$$ and joins $$\vee$$ as convex sums (convex hull of unions). In this case the minimal element of $$\mathsf{CONV}(V)$$ is empty set $$\emptyset$$, the minimal element of $$\mathsf{CONE}(V)$$ is $$\{0\}$$, and the maximal element in both lattices is $$V$$ itself.

For a convex set $$C\subset V$$ denote it support function as $$h_C : V^* \to \hat{\mathbb{R}}$$, where $$V^*$$ is a dual of $$V$$. That is, $$h_C(f) = \sup_{c \in C} f(c)$$. By dual cone of $$C$$ I mean a set $$C^\vee = \{ (t,f) \in \mathbb{R} \times V^* | h_C(f) \le t \} \in \mathsf{CONE(\mathbb{R} \times V^*)}$$

For Example, $$h_\emptyset(f) = - \infty$$, so $$\emptyset^\vee= \mathbb{R} \times V^*$$, and $$h_V(0) = 0$$, otherwise $$h_V(f) = + \infty$$, so $$V^\vee = \mathbb{R}_+ \times \{0\}$$.

As $$X \subset Y \Rightarrow \sup_{x \in X} f(x) \le \sup_{y \in Y} f(y)$$, it is obvious that the mapping $$(\bullet)^\vee : \mathsf{CONV}(V) \to \frac{\mathsf{CONE}(\mathbb{R} \times V^*)}{(\mathbb{R}_+ \times \{0\})}$$ is antitone, that is, it reverses the inclusion of order. Here I quatient $$\mathbb{R}_+ \times \{0\}$$ to make it a minimal element of the lattice. The mapping would be an antitone morphism if it reverses order and maps meets to joins, and joins to meets: $$(A \vee B)^\vee = A^\vee \wedge B^\vee \quad \& \quad (A \wedge B)^\vee = A^\vee \vee B^\vee \quad (1)$$

I think this result is true. For example, there is a well known result in the theory of dual cones that $$\bigcap^n_{i=1} C_i = \emptyset \iff \bigvee^n_{i=1} C_i^\vee = \mathbb{R} \times V^*.$$ But I have never seen this framed as a theorem in the form of (1), which seems to be a way more general statement. The theorem requers a proof. And I'm eager to write this proof. But at first, I decided to ask this question, to learn if this result is well known to be false, or some additional considerations need to be taken in account? Like topological properties of spaces, closednes of convex sets?