Finite Subset of Topological Vector Space is von Neumann-Bounded

Theorem

Let $\GF \in \set {\R, \C}$.

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

Let $F$ be a finite subset of $X$.

Then $F$ is von Neumann-bounded.

Proof

Let $F = \set {v_1, \ldots, v_m}$.

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

From Multiple of Vector in Topological Vector Space Converges, we have:

$\ds \frac {v_i} n \to {\mathbf 0}_X$ as $n \to \infty$

for each $1 \le i \le n$.

Hence for each $1 \le i \le n$, there exists $N_i \in \N$ such that:

$\ds \frac {v_i} n \in U$ for $n \ge N_i$

Let:

$N = \max \set {N_1, \ldots, N_m}$

Then for $n \ge N$ we have:

$v_i \in n U$ for each $1 \le i \le n$.

That is:

$F \subseteq n U$ for each $1 \le i \le n$.

Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $F$ is von Neumann-bounded.

$\blacksquare$