Can this set be open in a topological vector space?

Question

Let $X$ be a topological vector space, where the topology (as in Rudin's Functional Analysis) is such that,

• every singleton is a closed set,
• the vector space operations are continuous.

Now consider the set $S \subset X$ with the following property.

• $0 \in S$,
• for a nonzero $v \in X$ and all $0 < t < a \in \mathbb R$, we have $tv \notin S$.

Can $S$ be open ? some may argue that the open ball $B_r(0)$, an open beighborhood, intersects $tv$ and so $0$ is not an interior point of $S$ and therefore $S$ can't be open. But that is cheating, since there may be no metric or norm on the space to tell what is an open ball, I know that $0$ looks like a boundary point but how to show that for a vector topology. If my reasoning is wrong somewhere, please hint!

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Added: If am I allowed to take the discrete topology, which doesn't seem to be inconsistant with the vector topology above, then perhaps $S$ is open, but only if the field of scalars has this (discrete) topology too, otherwise, e.g. for $\mathbb R$, it is not going to be the case.

Correct me if I'm wrong

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Added: What are the further conditions on the vector topology such that $S$ is open, if we exclude the (trivial) discrete topology ?

Answer 1

Assume $S\subseteq X$ is open with $0\in S$. The map $\mu\colon \Bbb R\times X\to X$, $(c,w)\mapsto cw$ is continuous. The set $\mu^{-1}(S)$ is open and hence of the form $$\mu^{-1}(S)=\bigcup_{i\in I}\left]a_i,b_i\right[\times V_i$$ with real $a_i,b_i$ and open $V_i\subset X$. From $(0,v)\in\mu^{-1}(S)$, we see that there must exist some $i\in I$ with $a_i<0<b_i$ and $v\in V_i$. Pick $t\in\left]0,\min\{b_i,a\}\right[$. Then $tv\in S$.

Answer 2

Let $v$ be as in the second assumption about $S$. By continuity of the vector space operations, the function $f:\mathbb R\to X:t\mapsto tv$ is continuous. If $S$ were open then, since it contains $0=f(0)$, it would have to contain $f(t)$ for all $t$ sufficiently near $0$ in $\mathbb R$. That contradicts our choice of $v$, so $S$ is not open.