This topic is important because separation theorems can be used to derive properties of subdifferential calculus.
Recall: The projection $\bar{\omega} = \pi(\bar{x}; \Omega)$ is a singleton when Ω is closed and convex, and satisfies $\langle \bar{x} - \bar{\omega}, \omega - \bar{\omega} \rangle \leq 0$ for all $\omega \in \Omega$.
Proposition 1.1 (Strict Separation of a Point). Let Ω be a nonempty closed convex subset of ℝⁿ and let $\bar{x} \notin \Omega$. Then there exists $v \in \mathbb{R}^n$ (with $v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} < \langle v, \bar{x} \rangle.\]If this is true, we say that $\bar{x}$ and Ω are strictly separated.
Proof. Let $\bar{\omega} = \pi(\bar{x}; \Omega)$. Then $\langle \bar{x} - \bar{\omega}, \omega - \bar{\omega} \rangle \leq 0$ for all $\omega \in \Omega$.
Let $v = \bar{x} - \bar{\omega}$. Then
\[\langle v, \omega - \bar{\omega} \rangle \leq 0 \quad \forall \omega \in \Omega.\]So
\[\langle v, \omega - \bar{\omega} \rangle = \langle v, \omega - (\bar{x} - v) \rangle = \langle v, \omega \rangle - \langle v, \bar{x} - v \rangle = \langle v, \omega \rangle - \langle v, \bar{x} \rangle + \langle v, v \rangle \leq 0.\]Thus
\[\langle v, \omega \rangle + \|v\|^2 \leq \langle v, \bar{x} \rangle.\]Therefore
\[\sup\{\langle v, \omega \rangle \mid \omega \in \Omega\} < \langle v, \bar{x} \rangle. \quad \blacksquare\]Note: In the proof, $v \neq 0$ because $\bar{x} \notin \Omega$ while $\bar{\omega} \in \Omega$ (since Ω is closed).
Based on the above proposition, we have:
\[\sup\{\langle v, \omega \rangle \mid \omega \in \Omega\} < \langle v, \bar{\omega} \rangle.\]This means there exists $\alpha \in \mathbb{R}$ such that
\[\sup\{\langle v, \omega \rangle \mid \omega \in \Omega\} \leq \alpha < \langle v, \bar{\omega} \rangle.\]Definition 1.2 (Hyperplane and Half-Spaces). We can define the following sets:
\[L = \{x \in \mathbb{R}^n \mid \langle v, x \rangle = \alpha\}\] \[L^+ = \{x \in \mathbb{R}^n \mid \langle v, x \rangle > \alpha\}\] \[L^- = \{x \in \mathbb{R}^n \mid \langle v, x \rangle \leq \alpha\}\]Then we see $\Omega \subseteq L^-$ and $\bar{x} \in L^+$. So the hyperplane L separates $\bar{x}$ and Ω.
Theorem 2.1. Let $\Omega_1 \subseteq \mathbb{R}^n$ be nonempty, compact, and convex. Let $\Omega_2 \subseteq \mathbb{R}^n$ be nonempty and closed. Then there exists $v \in \mathbb{R}^n$ such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega_1\} < \inf\{\langle v, y \rangle \mid y \in \Omega_2\}.\]Lemma 2.2. In the setting of Theorem 2.1,
\[\Omega := \Omega_1 - \Omega_2\]is a closed subset of ℝⁿ.
(In other words, we want to show that for all $x \in \Omega_1$ and $y \in \Omega_2$, $\omega_0 = x - y$ satisfies $\omega_0 \in \Omega$.)
Proof. Let ${w_k}$ be a sequence in Ω that converges to $w_0$. We want to show $w_0 \in \Omega$. Then for all $k \in \mathbb{N}$, $w_k = x_k - y_k$. Since $\Omega_1$ is compact, the sequence ${x_k}$ contains a convergent subsequence ${x_{k_\ell}}$. Thus, we have $w_{k_\ell} = x_{k_\ell} - y_{k_\ell}$ and
\[y_{k_\ell} = x_{k_\ell} - w_{k_\ell} \to x_0 - w_0 \in \Omega_2\](because $\Omega_2$ is closed). Therefore $w_0 \in x_0 - \Omega_2 \subseteq \Omega_1 - \Omega_2$. Therefore Ω is closed. $\blacksquare$
Question: Under what conditions can $\Omega_1$ and $\Omega_2$ be separated by a hyperplane L?
Theorem 2.3 (Strict Separation of Disjoint Convex Sets). Let $\Omega_1$ be a nonempty compact convex set and let $\Omega_2$ be a nonempty closed convex set in ℝⁿ. Suppose that $\Omega_1 \cap \Omega_2 = \emptyset$. Then $\Omega_1$ and $\Omega_2$ can be separated strictly in the sense that there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega_1\} < \inf\{\langle v, y \rangle \mid y \in \Omega_2\}.\]Proof. Define $\Omega := \Omega_1 - \Omega_2$. By the previous Lemma, Ω is nonempty and closed. Furthermore, because $\Omega_1 \cap \Omega_2 = \emptyset$, we have $0 \notin \Omega$.
By the previous Proposition (strict separation of a point), there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\alpha := \sup\{\langle v, \omega \rangle \mid \omega \in \Omega\} < \langle v, 0 \rangle = 0.\]For any $x \in \Omega_1$ and $y \in \Omega_2$, we have $x - y \in \Omega$, so
\[\langle v, x - y \rangle \leq \alpha\] \[\langle v, x \rangle \leq \alpha + \langle v, y \rangle.\]This implies:
\[\sup\{\langle v, x \rangle \mid x \in \Omega_1\} \leq \alpha + \inf\{\langle v, y \rangle \mid y \in \Omega_2\}\] \[\Rightarrow \sup\{\langle v, x \rangle \mid x \in \Omega_1\} < \inf\{\langle v, y \rangle \mid y \in \Omega_2\}. \quad \blacksquare\]Define:
\[L = \{x \in \mathbb{R}^n \mid \langle v, x \rangle = \gamma\}\] \[L^+ = \{x \in \mathbb{R}^n \mid \langle v, x \rangle \geq \gamma\}\] \[L^- = \{x \in \mathbb{R}^n \mid \langle v, x \rangle \leq \gamma\}\]Then $\Omega_1 \subseteq L^-$ and $\Omega_2 \subseteq L^+$.
Review: Separating hyperplanes in Boyd et al. relate to ${x \mid x \in \mathbb{R}^n}$ and ${a^T x = \gamma \mid x \in \mathbb{R}}$.
Recall the property we proved: Let Ω be a nonempty closed convex set in ℝⁿ and let $\bar{x} \notin \Omega$. Then there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} < \langle v, \bar{x} \rangle.\]Remark (Lemma 2.4). Let Ω be a nonempty convex set and let $\bar{x} \notin \Omega$. Then there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} < \langle v, \bar{x} \rangle.\]Proof. Since Ω is convex, $\bar{\Omega}$ is also convex. Furthermore, $\bar{\Omega}$ is nonempty, closed, and convex. Since $\bar{x} \notin \bar{\Omega}$, there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \bar{\Omega}\} < \langle v, \bar{x} \rangle.\]Since $\Omega \subseteq \bar{\Omega}$, we have:
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} \leq \sup\{\langle v, x \rangle \mid x \in \bar{\Omega}\} < \langle v, \bar{x} \rangle. \quad \blacksquare\]Proposition 3.1. Let L be a subspace of ℝⁿ and let Ω be a nonempty subset of L with $\bar{x} \notin \bar{\Omega}$ and $\bar{x} \in L$. Then there exists $v \in L$ ($v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} < \langle v, \bar{x} \rangle.\]Definition 3.2 (Proper Separation). Let $\Omega_1, \Omega_2 \subseteq \mathbb{R}^n$ be nonempty and convex. We say that $\Omega_1$ and $\Omega_2$ can be properly separated if there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\langle v, x \rangle \leq \langle v, y \rangle \quad \forall x \in \Omega_1 \text{ and } \forall y \in \Omega_2 \quad (1)\]and there exists $\bar{x} \in \Omega_1$ and $\bar{y} \in \Omega_2$ such that
\[\langle v, \bar{x} \rangle < \langle v, \bar{y} \rangle. \quad (2)\]Note that (1) is equivalent to
\[\sup_{x \in \Omega_1} \langle v, x \rangle \leq \sup_{y \in \Omega_1} \langle v, y \rangle.\]The second property (2) can be written as follows:
\[\inf_{x \in \Omega_1} \langle v, x \rangle < \sup_{y \in \Omega_2} \langle v, y \rangle.\]Example. Consider $\Omega_1 = [a, b]$ and $\Omega_2$ = line containing $[a, b]$. What is v?
Definition 4.1 (Affine Set). A subset $A \subseteq \mathbb{R}^n$ is called an affine set if
\[\lambda a + (1 - \lambda)b \in A\]whenever $a, b \in A$ and $\lambda \in \mathbb{R}$.
Geometrically: A contains the line connecting any two points in A.
Examples:
Definition 4.2 (Affine Hull). Let $\Omega \subseteq \mathbb{R}^n$ be an arbitrary set. The affine set generated by Ω is the intersection of all affine sets containing Ω. We call this set the affine hull of Ω:
\[\text{aff}(\Omega) = \bigcap\{A \mid A \text{ is affine}, \Omega \subseteq A\}\]Proposition 4.3 (Properties of Affine Sets). Let $\Omega$ be an arbitrary set in ℝⁿ. Then we have the following:
(i) $\text{aff}(\Omega)$ is the smallest affine set that contains Ω.
(ii) $\text{aff}(\Omega) = \left{\sum_{i=1}^{m} \lambda_i \omega_i \mid \omega_i \in \Omega, \sum_{i=1}^{m} \lambda_i = 1, m \in \mathbb{N}\right}$
Proof. See book of Starck. $\blacksquare$
Definition 4.4 (Relative Interior). Let Ω be a convex set in ℝⁿ. An element $\bar{x}$ is said to be in the relative interior of Ω, denoted by $\text{ri}(\Omega)$, if $\bar{x}$ is the interior of Ω relative to $\text{aff}(\Omega)$.
That means there exists $\delta > 0$ such that
\[B(\bar{x}; \delta) \cap \text{aff}(\Omega) \subseteq \Omega.\]Example. Based on the above example, $\bar{x} \in \text{ri}(\Omega)$, but $a, b \notin \text{ri}(\Omega)$.
In other words, $\text{ri}(\Omega) = (a, b)$.
These properties are important for proving the proper separation property.
Theorem 4.5 (Properties of Relative Interior). Let Ω be a nonempty convex set in ℝⁿ. Then
(i) $\text{ri}(\Omega)$ is always nonempty. (Why?)
(ii) If $a \in \text{ri}(\Omega)$ and $b \in \bar{\Omega}$, then
\[\lambda a + (1 - \lambda)b \in \text{ri}(\Omega) \quad \text{if } 0 < \lambda \leq 1.\]Proof. Omitted. $\blacksquare$
Theorem 4.6 (Relative Interior of Minkowski Difference). If $\Omega_1$ and $\Omega_2$ are nonempty convex sets in ℝⁿ, then
\[\text{ri}(\Omega_1 - \Omega_2) = \text{ri}(\Omega_1) - \text{ri}(\Omega_2).\]Review this. Note that $\Omega_1 - \Omega_2 := {x - y \mid x \in \Omega_1, y \in \Omega_2}$. $\blacksquare$
Remarks:
(i) If $A \subseteq \mathbb{R}^n$ is affine and $0 \in A$, then A is a subspace of ℝⁿ.
Proof Sketch: The idea is we want to show:
Let $\alpha \in \mathbb{R}^n$, $r \in \mathbb{R}$. Then $r\alpha = r\alpha + (1 - r) \cdot 0$.
Let $x, y \in \mathbb{R}^n$. Then $x + y = \frac{1}{2}(2x + 2y) \in A$ (by affine scalar multiplication).
(ii) Any affine set $A \subseteq \mathbb{R}^n$ is a closed set. In particular, $\text{aff}(\Omega)$ is closed for any subset Ω of ℝⁿ.
Proof Sketch: Fix $x_0 \in A$ and let
\[L = A - x_0 = \{a - x_0 \mid a \in A\}.\]L is an affine subspace of ℝⁿ.
\[A = L + x_0 \Rightarrow A \text{ is a subspace, and subspaces of ℝⁿ are closed. Why?}\]Property of separation:
\(\langle v, x \rangle \leq 0 \text{ for all } x \in \Omega\) \(\langle v, \bar{x} \rangle < 0 \text{ for some } \bar{x} \in \Omega\)
(iii) The sequence ${x_k}$, where $x_k = -\frac{1}{k}x_0$, belongs to $\text{aff}(\Omega) = L$.
\(0 \in \bar{\Omega}\) \(\Omega \subseteq \text{aff}(\Omega)\) \(\Rightarrow \bar{\Omega} \subseteq \overline{\text{aff}(\Omega)} = \text{aff}(\Omega)\) \(\Rightarrow \text{aff}(\Omega) \text{ is a subspace of } \mathbb{R}^n.\) \(x_k = -\frac{1}{k}x_0 \in L.\)
Lemma 5.1. Let Ω be a nonempty convex set in ℝⁿ such that $0 \in \bar{\Omega} \setminus \text{ri}(\Omega)$. Then there exists a sequence ${x_k}$ in ℝⁿ such that $x_k \notin \bar{\Omega}$ for all $k \in \mathbb{N}$ and $x_k \to 0$ as $k \to \infty$.
The proof is based on the properties of relative interiors of convex sets.
Some ideas:
Proof. Since Ω is nonempty, there exists $x_0 \in \text{ri}(\Omega)$. Then $-tx_0 \notin \bar{\Omega}$ for all $t > 0$. By contradiction, suppose that $-tx_0 \in \bar{\Omega}$ for some $t > 0$. Then
\[0 = \frac{t}{1+t}x_0 + \frac{1}{1+t}(-tx_0) \in \text{ri}(\Omega).\]This is a contradiction to the assumption $0 \notin \text{ri}(\Omega)$.
Now, let $x_k = -\frac{1}{k}x_0$. By the above, $x_k \notin \bar{\Omega}$ for all $k$ (we verified this). Since $x_k \to 0$ as $k \to \infty$, we are done. $\blacksquare$
Lemma 5.2. Let Ω be a nonempty convex set in ℝⁿ. If $0 \notin \text{ri}(\Omega)$, then Ω and ${0}$ can be properly separated. That means, there exists an element $v \in \mathbb{R}^n$ such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} \leq \langle v, 0 \rangle = 0\] \[\inf\{\langle v, x \rangle \mid x \in \Omega\} < \langle v, 0 \rangle = 0.\]Don’t understand this. (Note from the original)
Proof (Lemma 5.2). We consider two cases.
Case 1: Assume $0 \notin \bar{\Omega}$.
In this case, we proved earlier that 0 and Ω can be strictly separated (first lecture).
This means there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\sup\{\langle v, x \rangle \mid x \in \Omega\} < \langle v, 0 \rangle = 0.\]Case 2: Assume $0 \in \bar{\Omega}$.
In this case $0 \in \bar{\Omega} \setminus \text{ri}(\Omega)$.
By Lemma 5.1, there exists a sequence ${x_k} \subset L = \text{aff}(\Omega)$ such that $x_k \notin \bar{\Omega}$ for all $k \in \mathbb{N}$, and $x_k \to 0$ as $k \to \infty$.
Then, by the strict separation property of subspaces (L is a subspace), there exists $v_k \in L = \text{aff}(\Omega)$ ($v_k \neq 0$), and
\[\sup\{\langle v_k, x \rangle \mid x \in \Omega\} < \langle v_k, x_k \rangle.\]In particular
\[\langle v_k, x \rangle < \langle v_k, x_k \rangle\]for all $x \in \Omega$ and $k \in \mathbb{N}$.
Dividing both sides by the norm of $v_k$, we obtain:
\[\left\langle \frac{v_k}{\|v_k\|}, x \right\rangle < \left\langle \frac{v_k}{\|v_k\|}, x_k \right\rangle \quad (*)\]Now let $w_k = \frac{v_k}{|v_k|}$. Then $|w_k| = 1$ and WLOG, we can assume that $w_k \to v$, where $|v| = 1$ (because $w_k$ is bounded and has a convergent subsequence $= w_k$).
By passing to a limit in $(*)$, we see that
\[\langle v, x \rangle \leq 0 \quad \text{for all } x \in \Omega\] \[\left(|\langle w_k, x_k \rangle| \leq \|w_k\| \|x_k\| = \|x_k\| \to 0\right)\]Cauchy-Schwarz
Writing it out…
\[\left|\left\langle \frac{v_k}{\|v_k\|}, x \right\rangle\right| = |\langle w_k, x \rangle| < |\langle w_k, x_k \rangle| \leq \|w_k\| \|x_k\| \leq \|x_k\|\]Then, as $k \to \infty$ we have:
\[\forall \varepsilon > 0, \quad |\langle w_k, x \rangle| < \|x_k\|\] \[\Rightarrow |\langle v, x \rangle| < 0\]for all $x \in \Omega$.
It remains to show there exists $\bar{x} \in \Omega$ such that $\langle v, \bar{x} \rangle < 0$. (Why??)
Observe that $v \in L$ (because $\bar{E}_{w_k} \in L$, and L is a closed subspace of ℝⁿ).
By contradiction, suppose that $\langle v, x \rangle \geq 0$ for all $x \in \Omega$.
Then $\langle v, x \rangle = 0$ for all $x \in \Omega$. Since $v \in L = \text{aff}(\Omega)$, $v = \sum_i \lambda_i \omega_i$ where $\lambda_i \in \mathbb{R}$, $\sum_{i=1}^{n} \lambda_i = 1$, and $\omega_i \in \Omega$. Then $|v|^2 = \langle v, v \rangle > 0$.
\[\langle v, \sum \lambda_i \omega_i \rangle = \sum \lambda_i \langle v, \omega_i \rangle = 0.\]This is a contradiction. $\blacksquare$
Theorem 5.3 (Proper Separation Theorem). Let $\Omega_1$ and $\Omega_2$ be two nonempty convex sets in ℝⁿ. If
\[\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2) = \emptyset,\]then $\Omega_1$ and $\Omega_2$ can be properly separated.
Proof. (Proof is mainly based on Lemma 5.2)
Define $\Omega := \Omega_1 - \Omega_2 = {x - y \mid x \in \Omega_1, y \in \Omega_2}$.
Then Ω is a nonempty convex set. (How do we know nonemptiness?)
Since $\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2)$ can be properly separated, the set Ω and ${0}$ can be properly separated. By Lemma 5.2,
\[0 \notin \text{ri}(\Omega) = \text{ri}(\Omega_1) - \text{ri}(\Omega_2).\]Since $\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2) = \emptyset$, $0 \notin \text{ri}(\Omega) = \text{ri}(\Omega_1) - \text{ri}(\Omega_2)$.
By Lemma 5.1 (applied for Ω and 0), there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\(\langle v, z \rangle \leq 0 \text{ for all } z \in \Omega \quad (1)\) \(\langle v, \bar{z} \rangle < 0 \text{ for some } \bar{z} \in \Omega \quad (2)\)
Fix any $x \in \Omega_1$ and $y \in \Omega_2$. Then $z = x - y \in \Omega_1 - \Omega_2 = \Omega$. Thus
\(\langle v, z \rangle \leq 0\) \(\Rightarrow \langle v, x - y \rangle \leq 0\) \(\Rightarrow \langle v, x \rangle \leq \langle v, y \rangle.\)
Since $\bar{z} \in \Omega$, $\bar{z} = \bar{x} - \bar{y}$ where $\bar{x} \in \Omega_1$, $\bar{y} \in \Omega_2$. Then
\[\langle v, \bar{z} \rangle = \langle v, \bar{x} \rangle - \langle v, \bar{y} \rangle < 0. \quad \#\]It follows that
\[\langle v, \bar{x} \rangle < \langle v, \bar{y} \rangle.\]Therefore $\Omega_1$ and $\Omega_2$ can be properly separated. $\blacksquare$
In the following Lemma, we will prove the theorem for a special case: $\Omega_1$ is an arbitrary set and $\Omega_2$ is a singleton.
Lemma 5.3 (Prolonging Lemma). Let Ω be a nonempty convex set in ℝⁿ. Suppose that $\bar{x} \in \text{ri}(\Omega)$ and $\bar{y} \in \bar{\Omega}$. Then there exists $t > 0$ such that
\[\bar{x} + t(\bar{x} - \bar{y}) \in \Omega.\]Proof. Since $\bar{x} \in \text{ri}(\Omega)$, there exists $\delta > 0$ such that
\[B(\bar{x}; \delta) \cap \text{aff}(\Omega) \subseteq \Omega.\]We can choose $t > 0$ sufficiently small such that
\[\bar{x} + t(\bar{x} - \bar{y}) \in B(\bar{x}; \delta).\]We have the following:
\[\bar{x} + t(\bar{x} - \bar{y}) = (1 + t)\bar{x} + (-t)\bar{y} \in \text{aff}(\Omega).\]Therefore
\[\bar{x} + t(\bar{x} - \bar{y}) \in B(\bar{x}; \delta) \cap \text{aff}(\Omega) \subseteq \Omega. \quad \blacksquare\]Lemma 5.4. Let Ω be a nonempty convex set in ℝⁿ. If Ω and ${0}$ can be properly separated, then
\[0 \notin \text{ri}(\Omega) \quad \left(\text{ri}(\{0\}) \cap \text{ri}(\Omega) = \emptyset\right).\]Proof. By the definition, there exists $v \in \mathbb{R}^n$ ($v \neq 0$) such that
\[\langle v, x \rangle \leq \langle v, 0 \rangle = 0 \text{ for all } x \in \Omega,\]and there exists $\bar{x} \in \Omega$ such that
\[\langle v, \bar{x} \rangle < 0.\]We proceed by contradiction.
By contradiction, suppose that
\[0 \in \text{ri}(\Omega).\]By Lemma 5.3, there exists $t > 0$ such that
\[0 + t(0 - \bar{x}) = -t\bar{x} \in \Omega.\]Then $\langle v, -t\bar{x} \rangle \leq 0 \Rightarrow \langle v, \bar{x} \rangle \geq 0$.
But this is a contradiction. $\blacksquare$
Theorem 5.4 (Proper Separation - Converse). Let $\Omega_1$ and $\Omega_2$ be nonempty convex sets in ℝⁿ. If $\Omega_1$ and $\Omega_2$ can be properly separated, then
\[\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2) = \emptyset.\]Proof. Define $\Omega := \Omega_1 - \Omega_2$.
Then Ω is a nonempty convex set. Since $\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2)$ can be properly separated, the set Ω and ${0}$ can be properly separated. By Lemma 5.4,
\[0 \notin \text{ri}(\Omega) = \text{ri}(\Omega_1) - \text{ri}(\Omega_2).\]Thus
\[\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2) = \emptyset. \quad \blacksquare\]Theorem (Proper Separation - Complete Statement). Let $\Omega_1$ and $\Omega_2$ be two nonempty convex sets in ℝⁿ. Then $\Omega_1$ and $\Omega_2$ can be properly separated if and only if
\[\text{ri}(\Omega_1) \cap \text{ri}(\Omega_2) = \emptyset.\]Remark. If $\Omega_1, \Omega_2 \in \mathbb{R}^n$ are nonempty convex sets such that $\Omega_1 \cap \Omega_2 = \emptyset$, then $\Omega_1$ and $\Omega_2$ can be properly separated.
Proof idea:
\(\text{ri}(\Omega_1) \subseteq \Omega_1\) \(\text{ri}(\Omega_2) \subseteq \Omega_2\) \(\Rightarrow \text{ri}(\Omega_1) \cap \text{ri}(\Omega_2) = \emptyset\)
Example. Consider $\Omega_1 = {\bar{x}}$ (singleton) and $\Omega_2 = \Omega$. Note that the proof proceeds in several steps.