# Is the algebraic interior relatively open in a closed convex set?

I am struggling with proving or disproving the following:

Let $X$ be a locally convex space and let $C\subset X$ be closed convex with non-empty algebraic interior which we denote by $C^i$ (recall that $x\in C^i$ iff $\forall v\in X$, there is $t>0$ such that $x+tv\in C$). Now, is the following statement (S) true:

$(S):\ \forall x\in C^i$, there is a neighborhood $V$ of $x$ such that $V\cap C\subset C^i$.

In other words is $C^i$ (topologically) relatively open in $C$?

Personally I think that it is false, but I could not find a counter-example.

You're right that it's false, even in the nice special case of metric spaces. In that context, $x$ being in the algebraic interior of $C$ means that for all unit vectors $v$ there exists $t>0$ such that $x+tv\in C$, while being in the topological interior requires that there exist $t>0$ such that, for all unit vectors $v$, $x+tv\in C$. In other words, the topological interior condition is uniform over directions but the algebraic interior is not.

This observation leads readily to a counterexample; there's one in this spoiler block (hover to see):

The closed convex set $\bigcap_{n=1}^\infty \{x : \langle x,e_n\rangle \le \frac1n\}$ in $c_{00}$, where $e_n$ are the standard basis vectors, has the origin in its algebraic interior but not in its topological interior.

• @ Steven Tashchuk Your example shows the distinction between the algebraic and topological interiors. However (S) is about something different. So, how does your example relate to (S)? Thank you. Commented Nov 6, 2014 at 19:38
• $0\in C^i$ but there is no such neighbourhood $V$, since any $\varepsilon$-ball centred at $0$ includes points on the boundary of $C$, which are not in $C^i$.
– user21467
Commented Nov 6, 2014 at 21:09
• @ Steven Tashchuk I would like to check the details. So, please confirm that in your notation $c_{00}$ is the (non-closed) subspace of $c_0$ formed by eventually vanishing sequences (only a finite number of non-zero entries) with the $\ell^\infty$ norm. Also, a remark on the set $C$ in your example: the boundary of $C$ coincides with $C$ since it has an empty interior. Thanks Commented Nov 7, 2014 at 13:42

Completion to @ Steven Tashchuk counter-example.

Recall that $C:=\{x=(x_n)_{n\ge1}\in c_{00}\mid |x_n|\le\frac{1}{n}\}\subset c_{00}$ and the norm here is $\|x\|=\sup_{n\ge1}|x_n|$.

For every $x=(x_n)_{n\ge1}\in c_{00}$ denote by $N_x=\max\{n\ge1\mid x_n\neq0\}<\infty$

Then $0\in C^i$ because for every $v\in c_{00}$ there is $t=\min\{\frac{1}{n|v_n|}\mid v_n\neq0,\ n\le N_v\}$ such that, for every $n\ge1$, $|tv_n|\le \frac{1}{n}$, that is $tv\in C$.

For every $r>0$ there is $m\ge1$ such that $\frac{1}{m}<r$. Define $y=(y_n)_{n\ge1}\in C$ by $y_n=0$ if $n\neq m$ and $y_m=\frac{1}{m}$. Then $y\in B(0;r)$ and $y\not\in C^i$ since for every $v=(v_n)_{n\ge1}$ with $v_m>0$ and for every $t>0$, $y_m+tv_m>\frac{1}{m}$, that is $y+tv\not\in C$.