# Why convergence implies cauchy in topological vector space?

The following definition is from Janich's Topology book :

Definition (Topological Vector Space). A $\mathbb{R}$-Vector space $(E,\tau)$ with a topological space structure is called a Topological Vector Space of its topological and linear structure are compatible in the following sense :

$A1.$ The subtraction ($-:E^2\rightarrow E$) is continuous.

$A2.$ The multiplication by scalar $\mathbb{R}\times E\rightarrow E$ is continuous.

$A3.$ $(E,\tau)$ is Hausdorff.

Consider definition of "convergence of a sequence to some point", given here.

Again, Let the following definition from wikipedia :

Definition. $\{x_n\}$ is a Cauchy sequence if for each $V\in\tau$ containing $0$, there is some $N$ such that for each $m,n > N$ : $x_m-x_n\in V$.

Now I want to verify that $$\text{Every convergent sequence is cauchy}.$$ But it seems we don't have enough tools to prove.

Let's start ! Suppose $\{x_n\}\rightarrow x$ and let $V$ be a given open set containing $0$. By continuity of addition or subtraction, $V+x$ is a neighborhood of $x$. Therefore, there's a $N$ such that : $$\forall\:m,n>N\:: \left\{ \begin{array}{l}x_m\in V+x\\ x_n\in V+x \end{array}\right.\Longrightarrow \left\{ \begin{array}{l}x_m-x\in V\\ x_n-x\in V \end{array}\right.\tag{*}$$ But by idea of metric spaces, we must choose a "smaller" neighborhood $x$ in order to remove $x$ from the relations $(*)$.

• Try to start with:if $x_n$ converges to $x$ then $\forall n>N$ $x_n \in x_N + V$. So $x_m, x_n \in x_N+V$ where $V$ is a neighborhood, now what does this tell you. – user108539 Mar 12 '15 at 12:17
• How could you say that for some $N$ : $$\forall\;n>N \:: x_n\in x_N+V$$ ? – Fardad Pouran Mar 12 '15 at 13:37
• Definition of convergence – user108539 Mar 12 '15 at 13:44
• Can you go to detail ? – Fardad Pouran Mar 12 '15 at 13:51
• Can you check this lecture notes page 183 ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/TVS.pdf – user108539 Mar 12 '15 at 13:56

You have to consider the continuity of the map $f : E \times E \rightarrow E, (a,b) \mapsto a-b$ at the point $(x,x)$ and choose a neighbourhood $W$ of $x$ such that $f(W,W) \subset V$. Your 'smaller' neighborhood is $W$.