Suppose $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence and $A = \{ x_n : n \in \mathbb{N} \}$ not closed. Show that there exists $x \in X$ such that $x_n \longrightarrow x$.

Since $A$ is not closed: $$ \forall y\in A , \exists \varepsilon > 0 s.t. S(y,\varepsilon) \subset A $$ and since $x_n$ is Cauchy: $$ \forall \varepsilon > 0 ,\exists n_0 \in \mathbb{N} s.t. \forall n,m \geq n_0 : \rho (x_n,x_m) < \varepsilon $$ But I can't see a way to connect these two to prove convergence. Some starting hints would be greatly appreciated.

  • $\begingroup$ What is $X$ ? Where the $x_n$ are from ? $\endgroup$ – Fred May 14 at 9:35
  • $\begingroup$ ... and what's $S(y,\varepsilon)$? $\endgroup$ – Saucy O'Path May 14 at 9:35
  • $\begingroup$ $X$ is a metric space and $S(y, \varepsilon)$ is an open sphere with center $y$ and radius $\varepsilon$. $\endgroup$ – LoneBone May 14 at 9:38

Take $x\in\overline{\{x_n\,|\,n\in\mathbb N\}}\setminus\{x_n\,|\,n\in\mathbb N\}$. Then there is some sequence of terms of the sequence $(x_n)_{n\in\mathbb N}$ converging to $x$. In other words (since $x$ itself does not belong to the sequence), $x$ is the limit of some subsequence of the sequence $(x_n)_{n\in\mathbb N}$. But whenever a Cauchy sequence has a convergent subsequence, the whole sequence converges to the limit of that subsequence.

  • $\begingroup$ +1. A quick answer. $\endgroup$ – Qurultay May 14 at 9:41
  • $\begingroup$ Why the fact that $x$ does not belong to the sequence ensures that it's the limit of some subsequence of $x_n$? $\endgroup$ – LoneBone May 14 at 9:42
  • $\begingroup$ There is a sequence $(n_k)_{k\in\mathbb N}$ such that $\lim_{k\to\infty}x_{n_k}=x$. Is it possible that the set $\{n_k\,|\,k\in\mathbb N\}$ is finite? No, because then there would be a $N\in\mathbb N$ such that $n_k=N$ infinitely many times and we would have to have $x_N=x$. But $x$ does not belong to the sequence. $\endgroup$ – José Carlos Santos May 14 at 9:45

Not closed means there is a sequence from the set which converges to a point not in the set. Let $(y_n)$ be a sequence from $\{x_n:n \geq 1\}$ such that $y_n \to y$ and $y\neq x_n$ for any $n$. For each $n$ we can write $y_n=x_{k_n}$ and this sequence cannot be equal to $x$ after some stage because $y\neq x_n$ for any $n$. Hence we can find a subsequence of $\{x_n\}$ which is convergent. It is well known fact that if a subsequence of a Cauchy sequence converges then the whole sequence converges.


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.