Let $V$ be a real vector space. Given a subset $A \subseteq V$, say that a point $x \in A$ lies in the algebraic interior of $A$ if every affine line $\ell$ that passes through $x$ has the property that $x \in (\ell \cap B)^\circ$. Here $\ell \cap B$ is a subinterval of $\ell \cong \mathbb{R}$, so we defined its interior $(\ell \cap B)^\circ$ by equipping $\ell$ with the Euclidean topology.

Say that a subset $U \subseteq V$ is algebraically open if the algebraic interior of $U$ is $U$ itself. We thereby get a topology on $V$.

Any affine functional $\pi: V \to \mathbb{R}$ is continuous with respect to this topology. Is this topology defined by the algebraically open subsets the coarsest topology such that every affine functional on $V$ is continuous map?

I had asked this question on Math Stack Exchange earlier.


1 Answer 1


This is not true, already in $\mathbf R^2$. Indeed, if $V$ is finite-dimensional, then the Euclidean topology is the coarsest topology for which all linear¹ maps $V \to \mathbf R$ are continuous: choosing a basis $e_1,\ldots,e_n$ and taking the corresponding coordinate projections $\pi_i \colon V \to \mathbf R$ shows that all boxes $(a_1,b_1) \times \cdots \times (a_n,b_n)$ have to be open, and these generate the Euclidean topology.

So it suffices to find a set that is open in your sense but not in the Euclidean topology.

Example. Let $V = \mathbf R^2$, set $v_0 = (1,0)$, and inductively choose points $v_1, v_2, \ldots$ such that $\lVert v_i \rVert = 2^{-i}$ and $v_i$ does not lie on the (finitely many) lines $\overline{v_jv_k}$ for $j,k < i$. Define $Z = \{v_0,v_1,\ldots\}$ and $U = V \setminus Z$. Then $U$ is not open in the Euclidean topology because the $v_i$ converge to $0 \not \in Z$. But for every line $\ell \subseteq V$, the intersection $\ell \cap Z$ contains at most $2$ points by construction, so $\ell \cap U$ is open.

(For the algebraic geometers in the room, this construction is remarkably similar to an example I laid out in this answer: being closed cannot be checked on each curve.)

¹It doesn't matter if you say "linear" or "affine linear" here, because any affine linear map differs from a linear one by a translation, and translation $\mathbf R \to \mathbf R$ is continuous.

  • $\begingroup$ Thank you for your construction. I realize that the definition a set being "algebraically open" applies when the set is first assumed to be convex. $\endgroup$
    – Colin Tan
    Jan 25, 2023 at 13:22

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