Show there must be another "partial limit"

Question

Let $P$ the set of all partial limits of the sequence $\{a_n\}$ (partial limit = limit of a subsequence).

It's given that $\{0,2\} \subseteq P$ and $\forall n\in \mathbb{N}. \left| {a_{n+1}-a_n} \right| < 1$.

Show that there must be another partial limit.

My try:
Assuming by contradiction there are only $2$ partial limits; $\{0,2\}$.
We have:
$$\eqalign{ & \forall \varepsilon > 0\exists K \in N.\forall k > K:\left| {{a_{{n_k}}} - 0} \right| < \varepsilon \cr & \forall \varepsilon > 0\exists L \in N.\forall l > L:\left| {{a_{{n_l}}} - 2} \right| < \varepsilon \cr} $$

Now, there must $k_0$ such that $k_0+1=l_0$.
From the two inequalities above we have:
$$2 - 2\varepsilon < \left| {{a_{{k_0}}} - {a_{{l_0}}}} \right| < 2 + 2\varepsilon $$

Since we can choose $\varepsilon$ to be as small as we want, clearly $$\left| {{a_{{k_0}}} - {a_{{l_0}}}} \right| > 1$$

Which is a contradiction to $$\left| {{a_{n + 1}} - {a_n}} \right| < 1$$

Am I right? I'm not 100% percent sure of what I did here.

Answer 1

You just need to throw in that you can choose $k_0$ to be arbitrarily large where the two subsequences alternate, otherwise your proof looks fine. Also instead of $k_0 = l_0 + 1$ it should be $n_{k_0} = n_{l_0} + 1$.