# Adherent point (convergence of subsequences)

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)

• 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 $x_{2n+1}$. – Andrés E. Caicedo Oct 8 '13 at 16:07
• @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? – Giiovanna Oct 8 '13 at 16:10
• 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. – Giiovanna Oct 8 '13 at 16:13

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.
• And why we need the hypotesis that $x_n$ is bounded? – Giiovanna Oct 8 '13 at 16:23
• 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.) – Andrés E. Caicedo Oct 8 '13 at 16:26