Let $\{ a_n \}_{n=1}^{\infty}$ be a sequence and $A=\{a|\exists a_{n_j}\rightarrow a\}.$ Suppose there exists $\{ x_n \}_{n=1}^{\infty}\subseteq A \ s.t.x_n \rightarrow a_0$. Prove $a_0\in A.$

I tried taking subsequence of $x_n$ which converges to $a_0$, and a subsequence of $a_n$ which converges to the subsequence of $x_n$ and do some algebraic manipulations, but got stuck.

Any help appreciated.

  • $\begingroup$ Find $a_{n_k}$ with $|a_{n_k} - a_0| < 1/k$. Hint: triangular inequality. $\endgroup$ – user251257 Mar 2 '17 at 19:54
  • $\begingroup$ What you are proving is that the set of subsequential limits is closed. To do this, consider $\{ x_{n, m}\}$ where $x_{n,m} \rightarrow x_n$ as $m\rightarrow\infty$. $\endgroup$ – i707107 Mar 2 '17 at 19:55
  • $\begingroup$ @i707107 What do you mean by $x_{n,m}$? $\endgroup$ – Itay4 Mar 2 '17 at 19:58
  • $\begingroup$ Because $x_n\in A$, there is a subsequence of $\{a_n\}$ that converges to $x_n$. $\endgroup$ – i707107 Mar 2 '17 at 20:00
  • $\begingroup$ @i707107 Yes, that's what I said I tried. $\endgroup$ – Itay4 Mar 2 '17 at 20:03

Fix $\varepsilon>0$.

Since $x_n\rightarrow a_0$, we have

$$\text{For}\ \,\frac\varepsilon2>0,\ \exists N\in\mathbb N\ \text{such that}\ \forall n\geq N,\ \text{we have}\ |x_n-a_0|<\frac\varepsilon2$$

Because $x_n\in A$, there exists a sequence $\{a_{n_j}\}$ with $a_{n_j}\rightarrow x_n$, and thus

$$\text{For}\ \,\frac\varepsilon2>0,\ \exists J\in\mathbb N\ \text{such that}\ \forall j\geq J,\ \text{we have}\ |a_{n_j}-x_n|<\frac\varepsilon2$$

Now by triangle inequality, $$|a_{n_j}-a_0|\leq|a_{n_j}-x_n|+|x_n-a_0|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$$

and this shows that $a_{n_j}\rightarrow a_0$, hence $a_0\in A$

This is not a formal proof but an idea of how to use the triangle inequality.


Since you are not really saying where any of these elements live, I will assume a metric space, but you can do a fairly similar argument by showing $a_0$ is an accumulation point of $x_n$ which are accumulation points of $\{a_n \}$. Note that for each $n$, there exists a subsequence $a_{n_j} \to x_n.$ Now, for $x_n$, consider the ball of radius $\frac{1}{n}$ centered at $x_n$, there exists $j_n$ so that for all $m \geq j_n$, $a_{n_m} \in B(x_n, \frac{1}{n}).$

Then consider the sequence $\{a_{j_n}\}_{n=1}^{\infty}.$ You can show that this sequence converges to $a_0$. For all $\epsilon > 0,$ there exist $n$ so that $d(x_n,a_0) > \frac{\epsilon}{2}$ (by convergence of $x_n$) and $d(x_n,a_{j_n}) > \frac{\epsilon}{2}$ (by construction of the sequence). Hence you have a subsequence of $\{a_n\}$ converging to $a_0$.

  • $\begingroup$ Sorry, but I'm not familiar with metric spaces. Could you please explain shortly how to prove with an accumulation point? $\endgroup$ – Itay4 Mar 2 '17 at 20:05
  • $\begingroup$ Recall that a point $x$ is an accumulation point of a sequence $\{a_n\}$ if for all open set $\mathcal{U}$ containing $x$, there are infinitely many $m$ so that $a_m \in \mathcal{U}.$ We know that $a_0$ is an accumulation point of $\{x_n\}$, in particular, there exist infinitely many $x_n$ in every such $\mathcal{U}$. Moreover, since $\mathcal{U}$ is open, there exists a ball centered at $x_n$ contained in $\mathcal{U}$. That balls contains infinitely many $a_{n_j}$. $\endgroup$ – Maxime Scott Mar 2 '17 at 20:09

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