Strongly Minkowski equivalence

Question

Assume that $(X, \{ p_i \}_{i \in I})$ is a locally convex space. $A,B \subset X$ are said to be strongly Minkowski separated iff there exists $j \in I$ and $z \in X$ such that one of the following two inequalities is satisfied: \begin{gather} \inf \{ p_j(a +z): a \in A \} > \sup \{ p_j(b + z): b \in B \} \\ \inf \{ p_j(b + z): b \in B \} > \sup \{ p_j(a + z): a \in A \} \end{gather} This definition is equivalent to the existence of $\delta > 0$ and $\lambda \in I$ such that: \begin{equation} p_\lambda(a - b) \geq \delta \hspace{.2cm} \forall a \in A, b \in B \hspace{2cm} [1] \end{equation} I can not get the prove of condition [1] implies the definition of Strongly Minkowski.

Answer 1

That implication seems false.

Take any normed space $E=P\oplus \mathbb{R}e$, set $A=P \oplus (1,\infty)e$, $B=P \oplus (-\infty,-1)e$.

The condition (1) is met. However, for any seminorm $p$ and any $z$, if either one of the strong separation inequalities holds, then $A$ or $B$ is bounded for $p$.

Now, since for all $x \in A$ (resp. $B$), $\lambda >1$, $\lambda x \in A$ (resp. $B$), it follows that $p(A)$ (resp. $p(B)$) is zero, and in both cases it follows that $p=0$, a contradiction.