# Is a single nontrivial convex set in a topological vector space enough to make it locally convex?

This is sort of a definition question.

While a tvs is locally bounded if it contains a bounded neighborhood of the origin, a tvs is called locally convex if it contains a fundamental system of neighborhoods of the origin, say $\{U_n\}_{n\in\mathbb{N}}$, so that each $U_n$ is convex. Why the difference between the two? Is there an example of a tvs which has a convex neighborhood of the origin but not a fundamental system of neighborhoods of the origin? Or is a single convex neighborhood sufficient to generate an entire neighborhood basis?

I'm also curious as to why Rolewicz in Metric Linear Spaces defines a tvs to be locally $p$-convex if it has a fundamental system of the neighborhoods each of whose moduli of concavity are at most $2^{1/p}$. Is that not the same as having a neighborhood basis of $p$-convex sets?

• The entire space is always a convex neighbourhood of $0$. And more generally, given any neighbourhood of $0$, its convex hull is a convex neighbourhood of $0$ (and that's not always the entire space if the space isn't locally convex). The difference to "bounded" is that everything that's smaller than a bounded set is also bounded, so once you have one bounded neighbourhood, you have a fundamental system of bounded neighbourhoods. For convex, you have no such implication, so you need to demand a fundamental system in the definition. – Daniel Fischer Aug 5 '13 at 18:06
• Right, I think that's exactly it, the subsets of bounded sets are bounded but subsets of convex sets aren't necessarily convex. Even if you scale it down it might not be suitably small, so to speak. – danzibr Aug 5 '13 at 18:16