Show that if $(a_n) \to \alpha$ then $(b_n) \to \alpha$.

Question

My question is as follows

Let $(b_n)$ be the sequence obtained from $(a_n)$ by deleting terms so that infinitely many remain. Show that if $(a_n) \to \alpha$ then $(b_n) \to \alpha$.

A previous question asks the same but defines $b_n = a_{n+1}$ and so is a "base case" of the above question. In that question, I simply stated that since $(a_n) \to \alpha$, we have that, given an $\varepsilon > 0$, there exists an $N \in \Bbb N$ such that

$$\lvert a_n - \alpha\rvert < \varepsilon,\ \forall n > N$$

and that if $b_n = a_{n+1}$ then we simply define $N^\prime = N - 1$ and have that, given $\varepsilon > 0$, there exists an $N'$, namely $N' = N-1$ such that

$$\lvert b_n - \alpha\rvert < \varepsilon,\ \forall n > N'$$

and hence have the result that $\lim b_n = \alpha$.

Given that the above is somewhat correct (hopefully), is it wise then for this question to define $b_n = a_{n+k}$ and simply set $M = N - k$ for a proof, or does this use an unnecessary assumption? (That is, that $b_n$ is a copy of $a_n$ shifted along $k$ spaces, rather than that we can simply remove some terms of $a_n$).

Answer 1

No, you need to allow deleting any set of terms, not just the first $k$, as long as infinitely many terms remain. But you can simply take the same $N$ that you had for the sequence $a_n$: if $n > N$, $b_n = a_m$ for some $m \ge n$, so $m > N$ and $|b_n - \alpha| = |a_m - \alpha| < \epsilon$.

Answer 2

Fix $\epsilon >0$. Since $(a_{n})\to\alpha$, there is an $N$ such that $|a_{n}-\alpha|<\epsilon$ provided $n\geq N$. Hence, it follows that $|b_{m}-\alpha|<\epsilon$ provided $m\geq N$, since $b_{m}=a_{j}$ for some $j\geq m\geq N$.

Answer 3

No - the assumption that $b_n = a_{n+k}$ is not valid, since you may be deleting every other term (e.g. if $a_n = \frac{1}{n}$, then $b_n = \frac{1}{2n}$. Then the tail ends of sequences will not be identical.

However, your statement for $a_n$:

$$\lvert a_n - \alpha\rvert < \varepsilon,\ \forall n > N$$

will still hold true for $b_n$ (removing terms in the sequence will not matter, since the statement holds for all $n > N$).