Proof of subsequence convergence

1. Nov 5, 2013

synkk

prove if $a_{2k} \rightarrow l$ and $a_{2k-1} \rightarrow l$ then $a_n \rightarrow l$ where $a_{2k}$ and $a_{2k-1}$ are subsequences of $a_n$

my attempt:

since: $a_{2k} \rightarrow l$ then $\forall \epsilon > 0$ $\exists N_1 \in \mathbb{R}$ s.t. $2k > N_1 \Rightarrow |a_{2k} - l|< \epsilon$

also, since $a_{2k-1} \rightarrow l$ then $\forall \epsilon > 0$ $\exists N_2 \in \mathbb{R}$ s.t. $2k -1 > N_2 \Rightarrow |a_{2k-1} - l | < \epsilon$

from the above we have $k > \dfrac{N_1}{2}$ and $k > \dfrac{N_2 + 1}{2}$ then let $N = max \{ \dfrac{N_1}{2}, \dfrac{N_2 + 1}{2} \}$ then for $n >N$ $|a_n - l | < \epsilon$

does this proof make sense, if not where could I improve?

2. Nov 5, 2013

LCKurtz

Your proof looks good. Personally, I prefer to begin such a proof with "Suppose $\epsilon > 0$" and then dispense with the $\forall$ statements by just using that $\epsilon$ in the argument.

3. Nov 5, 2013

synkk

thanks,

in my book they have done a different method: http://gyazo.com/6043e5a4f42915dac0f29b629c56f14c

going by their method wouldn't my method be incorrect and I could simply say let N be the maximum of N_1 and N_2?

4. Nov 5, 2013

LCKurtz

I think your proof is OK as it is. You have that if either $2k$ or $2k-1$ is greater than $N$ then $a_{2k}$ or $a_{2k-1}$ is within $\epsilon$ of the limit. If $n>N$ it fits one of those two cases.