0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – GEdgar Apr 19 '15 at 17:16
0
$\begingroup$

With GEdgar's hint:

(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 $d(x_{n_i}, n_{n_k}) < r^{-i}$.

(ii) If a Cauchy sequence has a convergent subsequence, then the whole sequence converges.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.