2
$\begingroup$

Definition: $a$ is said to be an adherent point (value) of a sequence $(x_n)$ if there exists a subsequence of $x_n$, denoted by $(x_{n_k})$ such that $\lim x_{n_k} = a$

Let $(x_n)$ be a bounded sequence. If $\lim a_n = a$ and $a_n$ is an adherent value of $x_n$, prove that $a$ is an adherent value of $x_n$.

My attempt of proof:

We need to show that there is a subsequence of $x_n$ such that it converges to $a$. Since all sequences have a monotonic subsequence, we can choose $x_{n_k}$ that is limited, by hypothesis, and monotonic. So, $x_{n_k}$ converges and, by definition, its limit is an adherent value. So we have:

$\forall \epsilon > 0, \exists k_1 \in \mathbb{N} $ such that $k \geq k_1 \Rightarrow |x_{n_k} - a_k| < \epsilon$

$\forall \epsilon > 0, \exists k_2 \in \mathbb{N}$ such that $ k \geq k_2 \Rightarrow |a_k - a| < \epsilon$

So, if we get an $N = \max${$k_1, k_2$} we will have both inequalities and by the triangle inequality (getting a particular $a_k$)

$| x_{n_k} - a| < 2 \epsilon$

Does that mean that $a$ is an adherent point of $x_n$? I'm not sure if that is correct (well, it looks like I am finding 2 different limits for the subsequence)

$\endgroup$
  • $\begingroup$ No, this does not work at all. The limit of the monotonic subsequence you denote $x_{n_k}$ does not need to have anything to do with the $a_n$ or $a$. For example, consider a sequence where $x_{2n}=0$ and $x_{2n+1}=1-1/2^n$. Then the sequence $x_{2n+1}$ is monotone and converges to $1$. However, $0$ is an adherent point of the sequence $x_n$, and there is no way to verify that just by looking at the monotone subsequence $x_{2n+1}$. $\endgroup$ – Andrés E. Caicedo Oct 8 '13 at 16:07
  • $\begingroup$ @AndresCaicedo do you have a suggestion to make this proof? I see why this not works: my $a_k$ is fixed, in the first inequality and it doesn't in the second one. So, should I take all the subsequences that converges to all the $a_n$'s? $\endgroup$ – Giiovanna Oct 8 '13 at 16:10
  • $\begingroup$ in this case, I don't see why we need the hypotesis "$x_n$ is bounded, since I know there are subsequences that converges to all the $a_n$'s using my hypotesis. $\endgroup$ – Giiovanna Oct 8 '13 at 16:13
1
$\begingroup$

As pointed out in my comment, your argument does not work, and cannot be fixed.

Here is one way to proceed. Since $a_1$ is an adherent point of the sequence $x_n$, there is an $n_1$ such that $|x_{n_1}-a_1|<1$. Having defined $x_{n_1},\dots,x_{n_k}$, choose $n_{k+1}$ so that $n_{k+1}>n_k$ and $|x_{n_{k+1}}-a_{k+1}|<1/(k+1)$. This is possible because $a_{k+1}$ is an adherent point of the sequence $x_n$. It is now easy to check that the subsequence $x_{n_k}$ so built is Cauchy, and converges to $a$.

In slightly more detail: $x_{n_{k+1}}$ exists since there is a subsequence of $x_n$ that converges to $a_{k+1}$. We just pick a term of the sequence, with index large enough to be beyond $n_k$, and large enough so the term is close enough to $a_{k+1}$.

That the subsequence so built converges to $a$ follows from the triangle inequality: $|x_{n_k}-a|\le|x_{n_k}-a_k|+|a_k-a|$, and both terms on the right are small enough if $k$ is sufficiently large.

$\endgroup$
  • $\begingroup$ And why we need the hypotesis that $x_n$ is bounded? $\endgroup$ – Giiovanna Oct 8 '13 at 16:23
  • $\begingroup$ We do not. (If your definition of adherent point allows the $a_n$ or $a$ to be one of $+\infty,-\infty$, the argument I gave needs a tiny adjustment.) $\endgroup$ – Andrés E. Caicedo Oct 8 '13 at 16:26

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.