Countable neighborhood base at zero in a topological vector space I'm trying to read a proof of the following theorem 

Here is my question: 
How can one show the existence of a neighborhood base with the following properties as indicated in the proof?


*

*$V_j\supset V_{j+1}$;

*$V_0=X$;

*$V_n+V_n\subset V_{n-1}$.

*$V_j$ is balanced



The first property can be obtained by the answer to the question Constructing a local nested base at a point in a first-countable space. And the third and fourth properies are related to the fact that for every nbhd $V$ of zero there exists a balanced nbhd $U$ of zero so that $U+U\subset V$. But I don't see how to put things together. 
 A: We choose an arbitrary neighbourhood base at zero, say $\{B_n\}_{n \in \mathbb N}$, and for each $n \in \mathbb N$ a balanced open set $O_n \subset B_n$ such that $0 \in O_n$. Then we define $V_0 := X$. If $V_n$ is already defined, we choose $U_n$ to be a balanced neighbourhood of $0$ such that $U_n + U_n \subseteq V_n$. Then we define
$$
V_{n+1} := V_n \cap O_n \cap U_n.
$$
Let's check all the properties: $V_0 = X$ and $V_n \supseteq V_{n+1}$ are clearly satisfied. Further, $V_{n+1} + V_{n+1} \subseteq U_n + U_n \subseteq V_n$. Now let's verify that we obtained a neighbourhood basis: Let $N$ be an arbitrary neighbourhood of zero. Since $\{O_n\}_{n \in \mathbb N}$ is a neighbourhood basis, there exists a $k \in \mathbb N$ such that $O_k \subseteq N$ (we can even obtain $B_k \subseteq N$). Then clearly $V_{k+1} \subseteq N$, and further, $V_{k+1}$ is a neighbourhood of zero as the intersection of neighbourhoods (pick open subsets $W_1 \subseteq V_n$, $W_2 \subseteq O_n$, $W_3 \subseteq U_n$ such that $0 \in W_1 \cap W_2 \cap W_3$, the set $W_1 \cap W_2 \cap W_3$ is open). Further, each $V_n$ is balanced as the intersection of balanced sets.
