Good morning everyone,
I know from the Bolzano Weierstrass Theorem that every bounded sequence has a convergent subsequence. However, I am trying to show that given a sequence {$a_{n}$}, if the subsequences {$a_{4n}$}, {$a_{4n+1}$}, {$a_{4n+2}$}, and {$a_{4n+3}$} converge to some limit L, then so does {$a_{n}$}. My reasoning so far has been,
|$a_{4n} - L$| $< e$ for all $4n \geq N_{1}$, |$a_{4n+1} - L$| $< e$ for all $4n+1 \geq N_{2}$, |$a_{4n+2} - L$| $< e$ for all $4n+2 \geq N_{3}$, and |$a_{4n+3} - L$| $< e$ for all $4n+3 \geq N_{4}$.
Let M = max{$N_{1}, N_{2}, N_{3}, N_{4}$}. Then for all $n \geq M$, |$a_{n} - L$| $\lt$ e. However, I feel as this if this is too straightforward, and was wondering if there's a reason this doesn't necessarily hold?
Alternatively, could we use Cauchy sequences in some way to show that the original sequence converges? i.e. somehow using that if $4n, 4n+1 \geq$ max{$N_{1}, N_{2}$}, |$a_{4n+1} - a_{4n}$| $\lt e$ for all $e \gt 0$?
Thank you in advance.
In my opinion we can a easier proof for this problem Let {$a_n$} $\subset R$ be a sequence such that all sequences....... {$a_{4n}$} ,{$a_{4n+1}$},{$a_{4n+2}$}and{$a_{4n+3}$} are converging sequence with limit $L$. We would like to show that $L$ is the limit of {$a_n$}. Suppose that $\alpha$ is an arbitrary positive number then there is a natural number $N$ such that$\vert L-a_{4n} \vert$ $\lt \alpha$,$\vert L-a_{4n+1} \vert$ $\lt \alpha$,|$L-a_{4n+2}$| $\lt\alpha$ and |$L-a_{4n+3}$|$\lt \alpha$(for each $n\ge N$). Since each $n \in N$ is the one of the form $4k,4k+1,4k+2,4k+3$ then we have |$L-a_n$|$\le$ |$L-a_{4n}$|+|$L-a_{4n+1}$|+|$L-a_{4n+2}$|+|$L-a_{4n+3}$| $\lt$ $4$ $\alpha$ . Since $\alpha$ is arbitrary then we have $\lim_{n\to infinity}$ $a_n$=$L$