Here are the definitions:
Let $X$ be a set. Another set $\mathcal C\subseteq \mathcal P(X)$ is called a convexity over $X$ if
- $\varnothing, X\in\mathcal C$
- $\mathcal C$ is closed under arbitrary intersections
- $\mathcal C$ is closed under arbitrary nested unions (that is, a union of a chain from $\mathcal C$ [w.r.t. subset inclusion] is in $\mathcal C$)
The pair $(X,\mathcal C)$ is called a convex space, and the elements of $\mathcal C$ are called the convex subsets of $X$.
Given two convex spaces $(X,\mathcal C_1)$ and $(Y,\mathcal C_2)$, a function $f:X\rightarrow Y$ is called convexity preserving, or simply convex, if for every convex set of $Y$ its preimage under $f$ is a convex subset of $X$.
The standard convexity over the real numbers is, unsurprisingly, the set of convex subsets. If the definition is required, a subset $S\subseteq\Bbb R$ is called convex if for all $x$ and $y\in S$ and any $\lambda\in[0,1]$ we also have that $\lambda x+(1-\lambda)y\in S$.
My question is based on the observation that a convex function over $\Bbb R$ is also continuous.
Abstractly, given two sets $X_1, X_2$; two topologies $\mathcal T_1, \mathcal T_2$; and two convexities $\mathcal C_1, \mathcal C_2$; there doesn't seem to be any reason that a convex function $f:(X_1,\mathcal C_1)\rightarrow (X_2,\mathcal C_2)$ should also be continuous as a function from $(X_1,\mathcal T_1)$ to $(X_2,\mathcal T_2)$. If this is the case, I'd expect that $\mathcal T_1$ and $\mathcal C_1$ satisfy some non-trivial relation (same for $\mathcal T_2$ and $\mathcal C_2$).
Is there a compatibility between the canonical topology and the canonical convexity of $\Bbb R$ which implies that all convex functions are continuous?
I don't really want a proof that convex implies continuous, unless you're using it to exemplify the possible relation.
Any help is appreciated, as always.