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.

  • $\begingroup$ @ 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. $\endgroup$ – Neutral Element Nov 6 '14 at 19:38
  • $\begingroup$ $0\in C^i$ 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 $C^i$. $\endgroup$ – user21467 Nov 6 '14 at 21:09
  • $\begingroup$ @ 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 $\endgroup$ – Neutral Element Nov 7 '14 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 $y+tv\not\in C$.


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