# Interior point in convex set on n.v.s

I'm reading Peter Lax's Functional analysis, and I have a question about a definition:

Definition. $$X$$ is a linear space over the reals, $$S$$ a subset of $$X$$. A point $$x_0$$ is called an interior point of $$S$$ if for any $$y$$ in $$X$$ there is an $$\epsilon$$, depending on $$y$$, such that $$x_0 + ty \in S \qquad\text{for all real t, |t|<\epsilon.}$$

I'm wondering if this definition of interior point is equivalent to the classical one (topological interior) in the case of a convex set. I was looking for a counterexample but I couldn't find it.

It is not the same thing. It is what is called the algebraic interior. The standard counterexample is below where the origin is not in the topological interior. • This set is not convex, so is not a counterexample. – alexp9 Jan 12 at 12:19
• @Rhcpy99 Oh, sorry, I missed the convexity requirement. Then take the space $c_{00}$ with the infinity norm and the convex set $S$ of all sequences $x\in c_{00}$ such that $|x_n|\le\frac{1}{n}$. The origin is in the algebraic interior, but not in the topological one. – A.Γ. Jan 12 at 13:33
• Thank you so much for explaining. I started wondering the same thing when I encountered the definition of the algebraic interior (which does not require a topology at all) in Kolmogorov and Fomin's book, as a lead up to the Minkowski functional and Hahn-Banach theorem. – avs Mar 11 at 16:36

EDIT (again) : what follows is 100% wrong and it is a good example of how NOT to write an answer.

EDIT: What's below is not true; it refers to the definition of "algebraic interior of a convex set", which is another thing. The definition in the original question is exactly the same as the usual one of interior point in the topological sense.

They're not equivalent. Consider a line segment in the plane. Its midpoint, or any point not in the extremes, is in the "algebraic interior" but not in the topological interior, since the latter is empty.

• But a line segment in the plane has no interior points as defined above. For example, consider the segment $S$ joining the origin and $(1,0)$. Take $x_0$ to be the midpoint, and $y = \langle 0, 1\rangle$. There is no value of $\varepsilon$ such that $x_0 + ty \in S$ for all $|t|<\varepsilon$---any such point will be off of the segment. – Xander Henderson Jan 11 at 13:59
• @XanderHenderson: Oh sorry, I read $y \in S$. That's a definition of algebraic interior that is used in convex analysis. Well, then this is just the ordinary definition of interior point in the topological sense. – Giuseppe Negro Jan 11 at 14:04
• Sorry, Giuseppe, it is not true after the edit. – A.Γ. Jan 11 at 16:34
• Well, ok, I'd better think more before answering next time. :-) – Giuseppe Negro Jan 11 at 16:43