# Rudin's functional analysis 3.18, every originally bounded subset of a locally convex space is weakly bounded.

Reading through the proof of the following

In a locally convex space $$X$$, every weakly bounded set is originally bounded, and viceversa

The trivial part of the proof.

Since every weak neighborhood of $$0$$ in $$X$$ is an original neighborhood of $$0$$...

As usually I get confused with these weak topologies...

Clearly by definition of weak topology every weak neighborhood of $$0$$ is an original neighborhood as well (more specifically it follows from the fact that the weak topology is coarsest, right)?

This trivial part continues with

... it is obvious from the definition of "bounded" that every originally bounded subset of $$X$$ is weakly bounded.

If $$A$$ is bounded then for every neighborhood $$V$$ of $$0$$ there's a constant $$t > 0$$ such that $$A \subset tV$$, therefore if $$A$$ is originally bounded, specifically we have for every weak neighborhood $$W$$ we have a constant $$t>0$$ such that $$A \subset tW$$.

The proof is as simple as this, right? It doesn't seem to me that local convexity is necessary.

You have proved only one way. It is true that any originally bounded set is weakly bounded even if the space is not locally convex. The converse is not true in general. A counterexample can be constructed in non-locally convex spaces as follows: let $$X=L^{p} (0,1)$$ with $$0 with the metric $$d(f,g)=\int_0^{1}|f(x)-g(x)|\, dx$$. it is known that the only continuous linear functional on this space is the zero functional!. Hence $$\emptyset$$ and $$X$$ are the only weakly open sets so the entire space $$X$$ is weakly bounded. It is obviously not not originally bounded.