# Open connected which does not contain open convex subsets

It is well known that convex implies connected, and it is clear that if $$X$$ is a locally convex topological vector space and $$\emptyset \neq A\subseteq X$$ is open then $$A$$ contains a nonempty open convex set.

Question. Does there exist a topological vector space $$X$$ and a nonempty open connected set $$A\subseteq X$$ such that $$A$$ does not contain any nonempty open convex subset?

• Isn't $X=L^p([0,1])$, with $0<p<1$ (with $A$ being the "unit ball") an example? I'm guessing any $X$ not locally convex should provide an example. Or at least any not locally convex $F$-space. – tomasz May 15 at 15:52
• Yes, $L^p([0,1])$ indeed works, since its unit ball is path-connected (we do not need to triangle inequality to hold but only the homogeneity of $\|\cdot\|_p$). Hence the claim works for all non-locally convex tvs for which the topology is induced by a quasi-norm. I don't know what happens with not-locally F-space, but it would be interesting to know. – Paolo Leonetti May 15 at 17:44
• @PaulFrost: My guess is that for topological vector spaces over the reals or the complex numbers, it shouldn't be possible. But my knowledge of non-normed vector spaces is cursory at best, so my guess could be very naive. – tomasz May 16 at 9:12

Yes. For instance, consider the space $$X=L^p([0,1])$$ for some $$p\in (0,1)$$, and take $$A:=\{x\in X\mid \lVert x\rVert_p<1 \}$$.
Then $$A$$ is open and path-connected (for every $$a\in A$$, $$t\mapsto ta$$ yields a path from $$0$$ to $$a$$), but it does not contain any nonempty convex open subset (the only convex open sets in $$X$$ are $$\emptyset$$ and $$X$$).