Let $a_n\subseteq\mathbb{R}$ be a sequence, where $a_n\to a$. Let $$A = \{a_1, a_2, a_3, \cdots\}.$$ Let $b_n \in A$. Prove that if $(b_n)$ converges, then $b_n \to a$ or $b_n \to a_N$, where $N \in \mathbb{N}$. Let $B = \{b_1, b_2, b_3, \cdots\}$.
I tried to do it in the following way (with a lot of help from the book "Analysis with an introduction to proof" by Steven Lay).
Suppose that $(b_n)$ is finite with k terms. and let $\{b_n\} = a_m, \cdots, a_n$. Since $(b_n)$ is finite, the minimum distance between two distinct points is positive. $(b_n)$ is also Cauchy, since convergent implies Cauchy. We know from the definition of Cauchy sequence, that for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $$s, t \geq N \implies | a_s - a_t | < \epsilon.$$ Given any $s\ge N$, $a_s$ and $a_N$ are both in $B$, so if the distance between them is less than $\epsilon$, it must be zero (since $\epsilon$ is the minimum distance between distinct points in S).
Thus $a_s$ = $a_N$ for all $s \ge N$. It follows that $$\lim_{n\to k} b_n = a_N.$$
Is this correct? And I'm struggling when we suppose that $B$ is infinite.
For the infinite case, there's a whole lot going in my mind. We know that B is a subset of $\mathbb{R}$, and B is bounded and infinite. The bolzano-weirstrass theorem then implies that there is a real number "x" in $\mathbb{R}$ which is a limit point of B and we know that B converges to x. We know $a_n$ converges to a, we just need to show that a = x. Showing x $\leq$ a has not been a problem but I am having difficulties in showing that x $\geq$ a. I got stuck here so I tried another method.
I tried to suppose that x > a and x < a give us contradictions but I can't come up with the contradictions.
I then tried another way, we know that B is infinite, so B = { $a_m$, $a_{m+1}$, .... }. Only a finite number of terms have been skipped, and so $ \lim (b_n) = a$. Is this correct..?
I haven't studied subsequences yet, so I can't use them.
