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I have recently come across this result.

Theorem: Let $A$ be a convex subset of a t.v.s. $X$. If $\operatorname{int} (A) \neq \emptyset$, then $\operatorname{aint} (A) = \operatorname{int} (A)$.

Here $\operatorname{aint} (A)$ is the algebraic interior of $A$.

My attempt: Clearly, $\operatorname{int} (A) \subset \operatorname{aint} (A)$. Let's prove the converse. Assume $a \in \operatorname{aint} (A)$. We want to show $a \in \operatorname{int} (A)$. Fix $b \in \operatorname{int} (A)$. There is $t>0$ such that $c := a+t(a-b) \in A$. It follows that $a = (1-\lambda)b+\lambda c$ with $\lambda := 1/ (1+t)$. Then $a \in U := (1-\lambda) \operatorname{int} (A) +\lambda c \subset A$. Notice that $(1-\lambda) \operatorname{int} (A)$ is open, so $U$ is open. This completes the proof.

  1. Could you have a check on my attempt?

  2. Is below related statement true? If not, please provide a counter-example.

Let $A$ be a convex subset of a t.v.s. $X$. If $\operatorname{aint} (A) \neq \emptyset$, then $\operatorname{aint} (A) = \operatorname{int} (A)$.

Analyst
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1 Answers1

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  1. Yes, this looks fine.

  2. If this assertion would be true, you would (together with 1.) always have $\operatorname{aint}(A) = \operatorname{int}(A)$.

It is well known, that this equality holds if $A$ is a closed, convex subset of a Banach space.

Here are some counterexamples:

  1. Let $X$ be an infinite-dimensional normed space and $\ell \colon X \to \mathbb R$ be linear and unbounded. Then $$ A = \{x \in X \mid |\ell(x)| < 1\} $$ has empty interior, but non-empty algebraic interior.

  2. Let $X = c_c$ be the space of finite sequences (equipped with the $\infty$-norm) and $$ A = \{x \in X \mid |x_n| \le 1/n \forall n \in \mathbb N\}. $$ Then, $A$ is convex and closed, has empty interior but non-empty algebraic interior.

gerw
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