# Are linear continuous mappings open on totally bounded sets?

Let $$X$$ and $$Y$$ be locally convex spaces, and $$\varphi: X\to Y$$ a linear continuous mapping. Suppose first that $$S$$ is a compact set in $$X$$. Then $$\varphi$$, being considered as a mapping from $$S$$ to $$\varphi(S)$$, $$\varphi\Big|_S:S\to \varphi(S)$$ is open in the sense that for any open set $$U$$ in $$S$$ (with respect to the topology induced from $$X$$) its image $$\varphi(U)$$ is an open set in $$\varphi(S)$$ (with respect to the topology induced from $$Y$$).

This is strange, I was sure that the same must be true if $$S$$ is not necessarily compact, but just totally bounded (since we can consider the extension of $$\varphi$$ to the completions of the spaces $$X$$ and $$Y$$), but recently I understood unexpectedly that I can't write an accurate proof. Does this mean that there is a counterexample?

So my question:

Is $$\varphi\Big|_S:S\to \varphi(S)$$ an open mapping for each totally bounded set $$S$$ in $$X$$?

Bounded sets, e.g., unit balls in normed spaces, are always totally bounded for weak topologies. So you only have to find such a ball with two incompatible weak topologies, which is easy to do—say the ball of a suitable dual space with the weak and the weak $$\ast$$ topology.