Let $X$ be a topological vector space. A family $x_i \in X, i\in I,$ is called a topologically Cauchy family if for every NH $U$ of zero, there exists $i_0\in I$ such that $$ i,j\geqslant i_0 \Rightarrow x_i-x_j \in U $$ If the topology on $X$ is induced by a metric $d$, then the family is called metrically Cauchy if for every $\varepsilon >0$, there exists $i_0\in I$ such that $$ i,j\geqslant i_0 \Rightarrow d(x_i,x_j)<\varepsilon. $$
Allegedly, neither implies the other in general, which I'm having a hard time understanding. The notions are equivalent if the metric is shift invariant, e.g a norm.
Suppose $\tau _X$ is induced by a metric $d$. Let $x_i\in X, i\in I,$ be metrically Cauchy. Let $B := B(0,\varepsilon)$ be a NH of zero. We can find $i_0\in I$ such that the metric condition is satisfied. But if the implication "Metrically Cauchy $\Rightarrow$ Topologically Cauchy" fails in general, that must mean $x_i-x_j\in B$ is not necessarily true for some $i,j\geqslant i_0$.
Conversely, if $x_i$ is topologically Cauchy and $\varepsilon >0$, then $x_i-x_j\in B(0,\varepsilon)$, but then again $d(x_i,x_j)< \varepsilon$ has to fail sometimes.
- What are some examples of non-shift-invariant metrics and families (sequences, even?) for which either implication fails?
- Is shift invariance necessary for the notions to be equivalent?
- Are there such topological vector spaces (whose topology is induced by a metric) such that exactly one implication is true?
Topologically Cauchy does not imply Metrically Cauchy.
Take an irrational number, say $e$. Define on $\mathbb Q$ $$ d(p,q) = \left\lvert \frac{1}{p-e} - \frac{1}{q-e} \right\rvert. $$ $d$ is not shift invariant. Note that $\tau _d$ is the subspace topology of the standard topology on $\mathbb R$. Put $$ x_n = \left ( 1+ \frac{1}{n} \right )^n, n\in\mathbb N $$ Then $x_n\to e$ in $\mathbb R$ so it is topologically Cauchy in $\mathbb Q$. On the other hand, given any $N\in\mathbb N$ one can find $m,n\geqslant N$ such that $d(x_m,x_n)$ is large.
For the converse, we can take in the same metric space $y_n = n,\ n\in\mathbb N$. Then $d(y_m,y_n)$ is eventually small, but clearly $y_n$ is not Topologically Cauchy.