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Let $X$ be a vector space, $Y$ a subspace of $X$, and $A \subset X$.

  • Then algebraic interior of $A$ with respect to $Y$ is defined as $$ \operatorname{aint}_{Y} (A) := \{x \in X \mid \forall y\in Y, \exists t >0: [x, x+t y] \subset A \}. $$ Clearly, $\operatorname{aint}_{Y} (A) \subset A$. If $A$ is convex, then the above definition reduces to $$ \operatorname{aint}_{Y} (A) = \{x \in X \mid \forall y\in Y, \exists t >0: x+t y \in A \}. $$

  • The algebraic interior of $A$ is defined as $$ \operatorname{aint} (A) := \operatorname{aint}_{X} (A). $$

Then I'm reading below result and its proof from this lecture note

Observation 0.7. Let $A$ be a subset of a vector space $X$ such that $\operatorname{aint} (A) \neq \emptyset$. Let $\alpha \in \mathbb R$ and $\ell: X \to \mathbb R$ be a nontrivial linear functional. Then the following statements are equivalent.

  • (i) $A \subset \{\ell \le \alpha\}$.
  • (ii) $\operatorname{aint} (A) \subset \{\ell \le \alpha\}$.
  • (iii) $\operatorname{aint} (A) \subset \{\ell < \alpha\}$.

Proof.

  • The implication (i) $\implies$ (ii) is obvious.

  • (ii) $\implies$ (iii) If (iii) is false, there exists $a \in \operatorname{aint} (A)$ such that $\ell(a)=\alpha$. Take a vector $v \in X$ such that $\ell(v)>0$. There exists $t>0$ with $a+t v \in A$, but then necessarily $a+\frac{t}{2} v \in \operatorname{aint} (A)$ and $\ell\left(a+\frac{t}{2} v\right)>\alpha$; thus (ii) is false.

  • (iii) $\implies$ (i) Assume that (iii) holds while (i) is false. Fix $a \in \operatorname{aint} (A)$ and $x \in A$ with $\ell(x)>\alpha$. Then the relatively open segment $(a, x)$ is contained in $\operatorname{aint} (A)$ and intersects the hyperplane $\ell^{-1}(\alpha)$, which contradicts (iii).

I suspect the the proof only holds with an additional assumption that $A$ is convex. Could you confirm if my understanding is correct?

Analyst
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1 Answers1

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You are correct. Without convexity, there is no reason to assume that the open segment $(a, x)$ must be contained in $A$, let alone $\operatorname{aint}(A)$.

As a concrete counterexample, let $$A = \{(x, y) \in \Bbb{R}^2 : x^2 + y^2 < 1\} \cup \{(2, 0)\}.$$ That is, $A$ is an open ball with an extra point $(2, 0)$. The algebraic interior of $A$ is the open ball, and is contained in the lower level set $\{\ell < 1\}$, where $\ell(x, y) = x$. But, $A \not\subseteq \{\ell \le 1\}$, as the right hand side does not contain $(2, 0)$.

Theo Bendit
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