Convergent Sequence in Topological Vector Space is Cauchy

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

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

Then $\sequence {x_n}_{n \in \N}$ is Cauchy.

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

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary, there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that:


 * $U + U \subseteq V$

Since $x_n \to x$, there exists $N \in \N$ such that:


 * $x_n \in x + U$

for $n \ge N$.

Since $U$ is symmetric, we have $U = -U$ and so:


 * $-x_m \in -x + U$

for $m \ge N$.

Then we have:


 * $x_n - x_m \in U + U \subseteq V$

for $n, m \ge N$.

So $\sequence {x_n}_{n \in \N}$ is Cauchy.