Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded/Proof 2

Theorem

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

Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence with $x_n \to x$.

Let:

$E = \set {x_n : n \in \N}$

Then $E$ is von Neumann-bounded.

Proof

From Union of Image of Convergent Sequence and Limit in Topological Space is Compact, $E \cup \set x$ is compact.

From Compact Subspace of Topological Vector Space is von Neumann-Bounded, $E \cup \set x$ is von Neumann bounded.

From Subset of von Neumann-Bounded Set is von Neumann-Bounded, $E$ is von Neumann bounded.

$\blacksquare$