If $\{x_n\}$ is a sequence s.t. $x_{2k} \rightarrow x$ and $x_{2k-1} \rightarrow x$, then $x_k \rightarrow x$
Just looking for feedback on my proof attemtpt
Proof Attempt
given that the odd and even terms of the sequence both individually converge that means: $$1) \ \exists \ N_1 \ s.t \ \forall \ k \leq N_1 \ |x_{2k} - x| < \frac{\epsilon}{2} \\ 2) \ \exists \ N_2 \ s.t \ \forall \ k \leq N_2 \ |x_{2k-1} - x| < \frac{\epsilon}{2} $$
If we choose $N = max\{N_1,N_2\}$
Then: $$|x_{2k} - x_{2k-1}| \leq |x_{2k} - x| + |x_{2k-1} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
I don't know if it is a big leap, but I'm assuming that $|x_{2k} - x_{2k-1}|$ represent consecutive sequential terms so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?
Ooooh. Ouch. No. That is not the right idea. I'd doesn't matter how close sequential terms get. The classic counter example is the harmonic series in which $a_n = \sum_{k = 1}^n \frac 1n$. $|x_n - x_{n-1}|=\frac 1n \to 0$ but $\{a_n\}$ does not converge.
In this case you are lucky in that you actually have and $x$ which $x_{2k}$ and $x_{2k - 1}$ converge to.
So for any $\epsilon$ you have an $N_1$ so that $n > N_1 \implies |x_{2n} - x| < \epsilon$ and you have an $N_2$ so that $n > N_2 \implies |x_{2n-1} - x| < \epsilon$.
So if you have $m > 2n > 2n-1$ where $n \ge \max (N_1, N_2)$ then if $m$ is odd then $m = 2k - 1$ for $k > n> N_2$ so $|x_m - x|= |x_{2k -1} - x| < \epsilon$. But if $m$ is even then $m = 2k$ for $k > n > N_1$ so $|x_m - x| = |x_{2k} - x| < \epsilon$.
In other words, let $M = 2\max (N_1, N_2)$. Then if $m > M$ then if $m = 2k$we have $k > N_1$ and if $m = 2k -1$ then we have $k > N_2$. ANd either way $|x_m - x| < \epsilon$.
.....
Now if you hadn't been given that $x_{2k}, x_{2k-1}\to x$ and where given that $x_{2k}$ and $x_{2k-1}$ were Chauchy and needed to prove $x_m$ was cauchy you would have had the right idea only you don't prove it only for the subsequent terms you must prove it for any TWO terms $m_1, m_2 > N$.
And we'd do this by taken cases.
Case 1: if $m_1, m_2$ are both even then $m_1, m_2 \ge N_1$ (um, why did you write $k \le N_1$? That was a typo I assume) so $|x_{m_1} - x_{m_2}| < \frac {\epsilon}2 < \epsilon.$
Case 2: if $m_1, m_2$ are both odd... some thing but with $N_2$.
Case 3: If $m_1$ is even , $m_2$ is odd are opposite parity then $|x_{m_1} - x_{m_2} \le |x_{m_1} - x_{2k}| + |x_{2k} - x_{2k -1}| + |x_{2k-1} - x_{m_2}| < \frac \epsilon 2 + \frac \epsilon 2 + \frac \epsilon 2 = \frac 32 \epsilon$.
So you'd have to modify for $\frac \epsilon 3$ instead.