Let $E$ be a Hausdorff TVS and let $(V_n)_{n \in \mathbb{N}}$ be a fundamental system of neighborhoods of $0$ in $E$ such that $\bigcap_{n \in \mathbb{N}} V_n=\{0\}$. Let us set $W_1=V_1$ and define by induction the sequence $(W_n)_{n \in \mathbb{N}}$ os balanced neighborhoods of $0$ which satisfy the relation $$W_{n+1}+W_{n+1}+W_{n+1} \subset V_n \cap W_n.$$ Let us define $\gamma$ on $E$ as follows:
- $\gamma(0)=0$
- $\gamma(x)=2^{-k}$ if $x \in W_k$ but $x \not\in W_{k+1}$
- $\gamma(x)=1$ if $x \not\in W_1$.
My question: How to prove that $x \in W_k$ if, and only if, $\gamma(x) \leq 2^{-k}$?
This is a statement in the proof of Theorem 5.10 of the Aliprantis, Border book Infinite Dimensional Analysis - A Hitchhiker’s Guide.
My attempt: If $x \in W_k$ then we have two cases to consider, $x \not\in W_{k+1}$ or $x \in W_{k+1}$. If $x \not\in W_{k+1}$ we obtain that $\gamma(x)=2^{-k}$ and we have the desired in(equality). In the same way, if $x \in W_{k+1}$ we have two cases to consider, $x \not\in W_{k+2}$ or $x \in W_{k+2}$. If $x \not\in W_{k+2}$ we obtain that $\gamma(x)=2^{-(k+1)}$, which implies that $\gamma(x) \leq 2^{-k}$. Assume that $x\neq 0$ since $E$ is Hausdorff, there exists a neighborhood $U$ of $0$ such that $U \cap \{x\}=\emptyset$. Since $\{W_k\}$ is a basis of neighborhoods, there exists $k_0 \in \mathbb{N}$ such that $x \not\in W_{k_0}$. Thus, the above process must stop. If $x=0$ clearly we have $\gamma(x)\leq 2^{-k}$ for all $k \in \mathbb{N}$. Therefore, $\gamma(x) \leq 2^{-k}$ whenever $x \in W_k$.
This answer has been moved from my comments to the question.
Hint
For the converse direction, if $\gamma(x) \leq 2^{-k}$, then we have $x = 0$ or $\gamma(x) = 2^{k_{0}}$ for some $k_{0} \geq k$. In this case, we have $x \in W_{k_{0}}$ but $x \not\in W_{k_{0} + 1}$ for some $k_{0} \geq k$.
As continued by @VictorHugo
Since $k_{0} \geq k$, then we have $W_{k_{0}} \subseteq W_{k}$. Hence, we have $x \in W_{k}$.