Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded

Theorem
Let $\struct {X, \tau}$ be a topological vector space.

Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence.

Let:


 * $E = \set {x_n : n \in \N}$

Then $E$ is von Neumann-bounded.

Proof
Let $W$ be an open neighborhood of ${\mathbf 0}_X$.

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, take a balanced open neighborhood of ${\mathbf 0_X}$, $V \subseteq W$.

From, there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that:


 * $U + U \subseteq V$

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, we can take $U$ to be balanced.

Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N \in \N$ such that:


 * $x_n \in x_N + U$

for $n \ge N$.

From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, there exists $M \in \N$ such that:


 * $x_N \in k U$

Then we have:


 * $x_n \in x_N + U \subseteq k U + U \subseteq k \paren {U + U} \subseteq k V$

for $n \ge N$.

Applying Topological Vector Space as Union of Dilations of Open Neighborhood of Origin again, for each $1 \le j \le N - 1$ we have $r_j \in \N$ such that $x_j \in r_j V$.

Taking $K = \max \set {r_1, \ldots, r_{N - 1}, k}$, we have:


 * $x_n \in K V$ for all $n \in \N$.

Since $V$ is balanced, we have that:


 * $x_n \in s V$ for all $s > K$.

That is:


 * $x_n \in s W$ for all $s > K$.