Is this invalid for the proof, $\{y_{k}\}$ converges if $\{y_{2k-1}\}$ and $\{y_{2k}\}$ converge to the same limit. In this question I know that lim$\{y_{2k-1}\}$ =lim$\{y_{2k}\}$= x, for some arbitrary constant. The problem is I don't know what operation are valid to show relation ship to $\{y_{k}\}$.
What I thought of saying  $\{y_{k}\}$=$\{y_{2k-1}\}$ $\cup$ $\{y_{2k}\}$ but this can't be used right? To my knowledge Union doesn't preserve ordering. Therefore the union of odd and even subsequence doesn't have to to be the same the sequence it creates.
Is there a procedure I can use to show these two sets combined form the other?
 A: Sequence is converged if it have only one limit point.
Let's consider any subsequence $y_{n_{k}}$. For sequence $n_{k}$ we have 3 variants:

*

*Only finite members of $n_{k}$ are odd

*Only finite members of $n_{k}$ are even

*$n_{k}$ have infinite members as odd, so even

In all cases we have for $y_{n_{k}}$ only one limit point $x=\lim\limits_{n \to \infty}y_{2n}=\lim\limits_{n \to \infty}y_{2n-1}$.
Addition:
Forecasting question for dividing $\mathbb{N}$ in $m$ subsuequences i.e. considering case $\mathbb{N}=N_1\cup \cdots \cup N_m$, where each $N_i$ infinite and disjoint for any pair. So suppose we have $m $ different in above meaning subsuequences and $\forall i, x=\lim\limits_{i \to \infty}y_{n_i}$. Then we can say, that $y_n$ will be converged i.e. each possible subsuequence will have same limit $x$.
But in case $m=\infty$ this statement fails: Let consider $2$ dimension infinite matrix filled with $0$-s and have $1$-s only on main diagonal. Then subsequence in each row will have limit $0$, while subsequence from diagonal have limit $1$.
A: Here's a proof using the $(\varepsilon,\delta)$ definition of a limit.
Since we are interested in the behavior of $\{y_k\}$ as $k\to\infty$, we may assume that $k>1$.
We're given that $\lim_{n\to\infty}y_{2k-1}=x$ and $\lim_{n\to\infty}y_{2k}=x$, so for all $\varepsilon >0$, there exist integers $N_1$ and $N_2$ such that $\left|y_{2k-1}-x\right|<\varepsilon$ for all $k>N_1$ and $\left|y_{2k}-x\right|<\varepsilon$ for all $k>N_2$. Note that for $k>1$, $2k>k$ and $2k-1>k$, so $2k>N_1$ and $2k-1>N_2$. This gives us sufficient conditions for when $\left|y_{2k-1}-x\right|<\varepsilon$ and $\left|y_{2k}-x\right|<\varepsilon$.
Let $N=\max\{N_1,N_2\}$. Given some integer $k>N$, we have that $k>N_1$ and $k>N_2$.
Now, $k$ is either even or odd. If it is odd, then $k=2m-1$ for some integer $m$, so our previous result allows us to write
$$k=2m-1>N_1$$
We can therefore conclude that $\left|y_k-x\right|=\left|y_{2m-1}-x\right|<\varepsilon$. A similar argument shows that $\left|y_k-x\right|<\varepsilon$ if $k$ is even. Since $k$ was given arbitrarily, the previous argument applies to all $k>N$. We have thus shown that for all $\varepsilon >0$, there exists an integer $N$ ($N=\max\{N_1,N_2\}$) such that $\left|y_k-x\right|<\varepsilon$ for all $k>N$. This proves that
$$\lim\limits_{k\to\infty}y_k=x$$
