Prove or disprove: The sequence has only got one cluster point if...

Question

Prove or disprove: The sequence $(x_{k})_{k\in\mathbb{N}}$ with $a_{k}=\frac{1}{k}$ if $k\in\mathbb{N}$ odd and $a_{k}=1$ if $k\in\mathbb{N}$ even, has only got one cluster point.

I say the statement is true because both the given sequences (odd / even) can only have 1 cluster point and so choosing one of these, we will end up with one cluster point.

Is it right or not?

Answer 1

The subsequence of odd $a_k$ has limit $0$, which is a cluster point of the of overall sequence $(a_k)$. The subsequence of even $a_k$ has limit $1$, which is a cluster point of the overall sequence $(a_k)$. We therefore conclude that the sequence $(a_k)$ has two cluster points: $0$ and $1$.

What we do not say is that "we can choose one of these cluster points to get a one cluster point, so the sequence has only one cluster point". There is no place in mathematics where one counts things in this way.

Answer 2

Disprove by considering two subsequences of {$a_n$}:
$\lim_{n\to\infty}${$a_{2n}$}$=1$ and
$\lim_{n\to\infty}${$a_{2n+1}$}$=0$.
Since the limit of these 2 subsequences are different, therefore {$a_n$} will not converge to one cluster point.