$ (a_n) $ is a sequence s.t. its partial limits are $ 3 , - 1 $. Define $ b_n = | a_n - 1 | $. Show $ b_n \rightarrow 2 $

Question

Problem: Let $ (a_n) $ be a sequence s.t. its partial limits are only from the set $ \{ 3 , - 1 \} $ ( infinite limits are not allowed to be partial limits of $ (a_n) $ ). Define a new sequence $ b_n = | a_n - 1 | $. Show that $ b_n \rightarrow 2 $.

Attempt: Suppose that $ b_n \nrightarrow 2 $. Meaning there exists $ \epsilon>0 $ s.t. $ \forall N \in \mathbb{N} . \exists n^*>N.| |a_n - 1 | - 2 | \geq \epsilon $. Choose $ N=1 $, then there exists $ n^* > N $ s.t. $ | |a_n^* - 1 | - 2 | \geq \epsilon \iff | |a_n^* - 1 | - 2 | \geq \epsilon \iff |a_n^* - 1 | - 2 \geq \epsilon$ or $ |a_n^* - 1 | - 2 \leq - \epsilon $.
[ I am stuck. I don't see how I can continue if I assume wlog either $ |a_n^* - 1 | - 2 \geq \epsilon $ or $ |a_n^* - 1 | - 2 \leq - \epsilon $, I think a better option would be If I wouldn't try to prove by contradiction, but that attempt was failure for me as well ].

Do you have any ideas how I should continue? or if I don't choose to prove by contradiction, how would one proceed after taking arbitrary $ \epsilon > 0 $ and using the givens that $3 ,-1 $ are partial limits?

Edit: The teacher-assistant told me that the theorem is true.

Answer 1

Note that since there are no infinite partial, or sub-sequential limits, then $\left(a_n\right)_{n\in\mathbb{N}}$ is bounded.

We claim that for all $\varepsilon>0$, there exists an $N\in\mathbb{N}$ such that for all $n\geq N$, we have $|a_n-(-1)|<\varepsilon$ or $|a_n-3|<\varepsilon$.

Suppose, for contradiction, that there exists an $\varepsilon>0$ such that for all $N\in\mathbb{N}$, there exists an $n\geq N$ with $|a_{n}-(-1)|\geq\varepsilon$ and $|a_{n}-3|\geq\varepsilon$. After possibly permuting or re-ordering terms, this gives rise to a sub-sequence $\left(a_{n_k}\right)_{k\in\mathbb{N}}$.

Since this sub-sequence $\left(a_{n_k}\right)_{k\in\mathbb{N}}$ is bounded, then by Bolzano-Wierstrass, there is a convergent sub-sequence of it, which is a partial limit of $\left(a_n\right)_{n\in\mathbb{N}}$. This new partial limit cannot be $-1$ or $3$, which is a contradiction, thus the claim holds.

Define the sequence $b_n=|a_n-1|$. We want to show that $\lim\limits_{n\to\infty}b_n=2$.

Let $\varepsilon>0$. Then there exists an $N\in\mathbb{N}$ such that for all $n\geq N$, we have $|a_n-(-1)|<\varepsilon$ or $|a_n-3|<\varepsilon$.

Let $n\geq N$. Now, there are two cases: $a_n-1\geq 0$ or $a_n-1<0$. In the former case, we have that $$|b_n-2|=||a_n-1|-2|=|a_n-3|$$ In the latter case, we have that $$|b_n-2|=||a_n-1|-2|=|-a_n-1|=|a_n-(-1)|$$ Now $|b_n-2|<\varepsilon$, thus $\lim\limits_{n\to\infty}b_n=2$.

Edit 2: To make the cases more clear, picking up from "We want to show that $\lim\limits_{n\to\infty}b_n=2$", this is equivalent to showing that for all $n\geq N$, we have $$||a_n-1|-2|<\varepsilon\Longleftrightarrow-\varepsilon<|a_n-1|-2<\varepsilon \Longleftrightarrow 2-\varepsilon < |a_n-1|<2+\varepsilon$$

In the case where $a_n-1\geq 0$, we have $$3-\varepsilon <a_n< 3+\varepsilon\Longrightarrow |a_n-3|<\varepsilon$$ The case when $a_n-1<0$ is similar.

Edit 1: I did not read the comments and I misinterpreted the statement of the question previously which lead to the answer below.

I think this statement is false. Consider the sequence $\left(a_n\right)_{n\in\mathbb{N}}$ defined as

$$\begin{cases} n &\text{ if }n\equiv 1\mod 3\\ -1&\text{ if }n\equiv 2\mod 3\\ 3&\text{ if }n\equiv 0\mod 3 \end{cases} $$

The only finite, partial limits of $\left(a_n\right)_{n\in\mathbb{N}}$ are $-1$ and $3$.

The sequence $b_n=|a_n-1|$ is seen to be $$b_n = \begin{cases} n-1 &\text{ if } n\equiv 1\mod 3\\ 2 &\text{ if else } \end{cases}$$ but $\left(b_{n}\right)_{n\in\mathbb{N}}$ does not converge, since $\limsup\limits_{n\to\infty}(b_n)=\infty$ and $\liminf\limits_{n\to\infty}(b_n)=2$.