# Is the following a valid characterisation of complete metric spaces?

A metric space $(X, d)$ is called complete if and only if every Cauchy sequence converges. Now does the following hold:

A metric space is complete if and only if every sequence $(x_i)_{i\in\mathbb N}$ where $d(x_j, x_k) < r^{-i}$ whenever $i \ge j,k$ converges to some point $x \in X$. (where $r > 1$ is a real number, for example $r = 2$)?

Of course, if a metric space is complete, then each such sequence converges, because it is in particular a Cauchy sequence, but the set of such sequences is not the set of all Cauchy sequences, as for example if $(x_i)$ is a Cauchy sequence and $\varepsilon := r^{-i}$ is given, then $N$ could be much larger then $i$, so that $d(x_j, x_k) < \varepsilon$ holds for all $j,k > N$, but not for all $i \le j,k \le N$:

So, is there an example of a non-complete metric space, so that every sequence $(x_i)$ such that $d(x_j, x_k) < r^{-i}$ for $j,k \ge i$ has a limit point in the space?

• Hints: (1) Show that a Cauchy sequence has a subsequence with the $r^{-i}$ property you mention. (2) If a Cauchy sequence has a convergent subsequence, then the whole sequence converges. – GEdgar Apr 19 '15 at 17:16

(i) Show that a Cauchy sequence has a subsequence with the $r^{-i}$ property.
Let $(x_n)$ be a Cauchy sequence, then for $i=1,2,3,\ldots$ there exists some minimal $N_i$ such that $d(x_j, x_k) < r^{-i}$ for $j,k \ge N_i$, set $n_{i} := N_i$ (by the minimality of the $N_i$ we have $N_1 \le N_2 \le N_3 \le \ldots$). Then the subsequence $(x_{n_i})$ has the property, because if $j,k \ge i$, then $n_j, n_k \ge n_i = N_i$, and so $d(x_{n_i}, n_{n_k}) < r^{-i}$.
Let $(x_n)$ be a Cauchy sequence and $(x_{n_k})$ be a convergent subsequence with limit point $x$. Let $\varepsilon > 0$, then there exists $N_1, N_2$ such that $$| x_j - x_i | < \varepsilon \quad \mbox{ and } \quad | x - x_{n_k} | < \varepsilon$$ for $i,j \ge N_1$ and $k \ge N_2$, set $N_3 := n_{N_2}$, then for $n_k \ge N_3$ we have $| x - x_{n_k} | < \varepsilon$. Set $N := \max\{N_1, N_3\}$, now for arbitrary $k \ge N$ $$| x - x_k | = | x - x_{n_k} + x_{n_k} - x_k | \le | x- x_{n_k} | + |x_{n_k} - x_k|$$ and as $n_k \ge k \ge N_3$ we have $|x - x_{n_k}| < \varepsilon$ and as $n_k \ge k \ge N_1$ also $|x_{n_k} - x_k| < \varepsilon$, taken together $$|x - x_k| < 2\varepsilon$$ which shows that $x_k \to x$.
So that each Cauchy sequences converges if every "$r^{-i}$"-sequences converges, showing that the space is complete.