How to define a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}$ such that each convergent sub-sequence has limit 1, but $(a_n)_{n\geq 0}$ doesn't converge? Part of a weekly submission assignment is about:
How to define a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}$ such that each convergent sub-sequence of $(a_n)_{n\geq 0}$ has limit 1, but $(a_n)_{n\geq 0}$ does not converge?
Given that $(V, d_v)$ is a metric space, $(a_n)_{n\geq 0}$ is a sequence in $V$ with $ a \in V$
However, me and my fellow students have been trying to figure it out for some time now, but we can't...
 A: Maybe this sequence can help you.
\begin{equation}
a_n=\left\{
\begin{array}{ccc}
1 & \text{if} & n \ \text{is even}\\
n & \text{if} & n \ \text{is odd}
\end{array}
\right.
\end{equation}
Observe that $a_n$ doesn't converges because is an unbounded sequence. Now, if $(a_{n_k})_{k\in\mathbb{N}}$ is a convergent subsequence to $x_0\in\mathbb{R}$ then we have the next cases

*

*If $x_0<1$ take $\varepsilon>0$ such that $(x_0-\varepsilon,x_0+\varepsilon)\subseteq (-\infty,1)$. All the values of the sequence are in $[1,\infty)$. Therefore is impossible that the subsequence converges to $x_0$.

*If $x_0>1$ take $\varepsilon>0$ such that $(x_0-\varepsilon, x_0+\varepsilon)\subseteq (1,\infty)$. Again, is impossible that the subsequence converges to $x_0$ because we can found an infinitely many terms in the subsequence that are greather than $x_0+\varepsilon$ or less than $x_0-\varepsilon$.

This proves that if $(a_{n_k})$ is a convergent subsequence, the only posibility is that $a_{n_{k}}$ converges to $x=1$.
A: $1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,...$
