For a vector topological space, a sequence $(x_n)$ is Cauchy when $\forall$ neighbourhood $U$ of $0$, there exists $N$ such that $m,n>N$ implies $x_m-x_n\in U$.

For a metric topological space $(E,d)$, a sequence $(x_n)$ is Cauchy when $\forall \epsilon>0$, there exists $N$ such that $m,n>N$ implies $d(x_m,x_n)<\epsilon$.

QUESTION : are these two definitions equivalent in the case of a metric topological vector space ?

If the distance is invariant by translation, I can prove it. But in general no.

QUESTION : if the previous question has answer "no", what is the "good" definition in the intersection ?


1 Answer 1


The answer is : yes if the vector topological space is metrizable, that is IFF it has a countable basis of topology on every point.

See theorem 3.21 here : http://ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/TVS.pdf (difficult)

And also the fact that a metrizable topological vector space has always a countable local basis of topology. This one is easy : if $A$ is closed in $V$, then consider the open sets $V_n=\{x\in X\,st\, d(A,x)<1/n\}$. We have $A=\bigcap_{n\geq 1}V_n$.

If you read French, I'll write the answer with some details here : http://laurent.claessens-donadello.eu/pdf/mazhe.pdf in the chapter "Topologie générale"


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