weave of two sequences Let x, y, z be three sequences of real numbers. z is said to be a weave (not "the weave", because weaves are not unique) of x and y iff x and y are disjoint subsequences of z that span all of z. For example, (0,0,1,0,1,0,0,1,0,1,...etc) is a weave of (0,0,0,...) and (1,1,1,..). My question is this: If x and y both converge to L, does any weave of them also converge to L? If so, give me a proof. If not, give me a counterexample.
 A: Suppose that $\mathbf{x}$ and $\mathbf{y}$ converge to $L$.  Then $\mathbf{x}$ and $\mathbf{y}$ are both Cauchy, and for each $\epsilon>0$ there exists $N$ such that $n,m>N$ implies that $|x_n-x_m|<\epsilon$.  A similar statement holds for $\mathbf{y}$.  Take any weave $\mathbf{z}$ of $\mathbf{x}$ and $\mathbf{y}$, and choose $\epsilon>0$.  Then choose $N_x'$ such that $|x_n-x_m|<\epsilon/2$ for $n,m>N_x'$, choose $N_x''$ such that $|x_n-L|<\epsilon$ for $n>N_x''$, define $N_x=\max\{N_x',N_x''\}$, and similarly for $\mathbf{y}$, giving $N_y$.  Now let $N$ be a natural number such that if $i$ is the index of $\mathbf{z}$ such that the $N_x$th entry of $\mathbf{x}$ appears and $j$ the index of $\mathbf{z}$ such that the $N_y$th entry of $\mathbf{y}$ appears, then $N>\max\{i,j\}$.  Now let $n,m>N$.  Then $|z_n-z_m|$ is either $|x_n-x_m|$, $|y_n-y_m|$, or $|x_n-y_m|$.  In the former two cases, these are clearly less than $\epsilon$.  In the latter case, we use the triangle inequality to give $|x_n-L+L-y_m|\leq |x_n-L|+|y_m-L|<\epsilon$.  Thus, $\mathbf{z}$ is Cauchy.  Since a Cauchy sequence converges to the same limit that any subsequence converges to, it converges to $L$.
A: If $x_n$ and $y_n \to L$, then there is some $N$ so that $|x_n - L| < \epsilon$, $|y_n - L| < \epsilon$ for $n>N$. So if $z_k$ is a sequence consisting of the $x_n$'s and $y_n$'s, there is some $K$ so that when $k>K$, $z_k = x_n$ or $z_k = y_n$ for some $n > N$.
So it seems like any weave of $x$ and $y$ will satisfy this. Certainly, the "trivial" weave, where you just alternate $x$ and $y$; here, you just need $K = 2N$.
A: Suppose $x_n$ and $y_n$ converge to $L$. Fix $\epsilon > 0$. Then we have some $N$ so that for $n \ge N$  both $\lvert x_n - L\rvert < \epsilon$ and $\lvert y_n - L \rvert < \epsilon$ hold. 
Now let $z_n$ be a weave of $x_n$ and $y_n$. 
After finitely many indices $z_n$ has used all of $x_1, \cdots, x_N$ and $y_1, \cdots, y_N$. After this point $\lvert z_n - L \rvert < \epsilon$ 
A: Suppose that $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$ converge to $L$. Then for each $\epsilon>0$ there is an $m(\epsilon)\in\Bbb N$ such that $|x_n-L|<\epsilon$ and $|y_n-L|<\epsilon$ whenever $n\ge m(\epsilon)$. Let $z=\langle z_n:n\in\Bbb N\rangle$ be any weave of $x$ and $y$. For each $\epsilon>0$ there is a $k(\epsilon)\in\Bbb N$ such that the first $m(\epsilon)$ terms of $x$ and the first $m(\epsilon)$ terms of $y$ all precede $z_{k(\epsilon)}$. It follows that $|z_n-L|<\epsilon$ whenever $n\ge k(\epsilon)$.
