Convergence of $\operatorname{mid}\{(x_n),(y_n),(z_n)\}$ where $(x_n)$, $(y_n)$ and $ (z_n)$ are convergent sequences If $(x_n)$, $(y_n)$ and $(z_n)$ are three convergent sequences then show that, the sequence $(w_n)$ defined by $(w_n)=\operatorname{mid}\{(x_n),(y_n),(z_n)\}$ is also convergent.
Does anyone have a proof for this? Please help.

Let $x_n\rightarrow x$,$y_n\rightarrow y$,$z_n\rightarrow z$ and suppose that $\varepsilon>0$ is arbitrary positive number. Then there exist $N_x,N_y,N_z\in\mathbb{N}$ such that $$\forall n\geq N_x, \forall n\geq N_y,\forall n\geq N_z,\quad |x_n-x|<\varepsilon,|y_n-y|<\varepsilon,|z_n-z|<\varepsilon$$
Set $N=\max\{N_x,N_y,N_z\}$. Now for any $n\geq N$ we have
$$|\frac{x_n+y_n+z_n}{3}-\frac{x+y+z}{3}|=|(\frac{x_n}{3}-\frac{x}{3})+(\frac{y_n}{3}-\frac{y}{3})+(\frac{z_n}{3}-\frac{z}{3})|\leq\\ |(\frac{x_n}{3}-\frac{x}{3})|+|(\frac{y_n}{3}-\frac{y}{3})|+|(\frac{z_n}{3}-\frac{z}{3})|\leq \frac{1}{3}|x_n-x|+\frac{1}{3}|y_n-y|+\frac{1}{3}|z_n-z|<\\ \frac{1}{3}\varepsilon+\frac{1}{3}\varepsilon+\frac{1}{3}\varepsilon=\varepsilon$$
Hence $\frac{x_n+y_n+z_n}{3}\rightarrow\frac{x+y+z}{3}$.
Edit:
$$\min\{\max\{x_n,y_n\},\max\{x_n,z_n\},\max\{z_n,y_n\}\}=\min\{\frac{x_n+y_n+|x_n-y_n|}{2},\frac{x_n+z_n+|x_n-z_n|}{2},\frac{z_n+y_n+|z_n-y_n|}{2}\}$$
and also if we set $\alpha=\frac{x_n+y_n+|x_n-y_n|}{2},\beta=\frac{x_n+z_n+|x_n-z_n|}{2},\gamma=\frac{z_n+y_n+|z_n-y_n|}{2}$ then
$$\min\{\alpha,\beta,\gamma\}=\min\{\min\{\alpha,\beta\},\gamma\}=\min\{\frac{\alpha +\beta-|\alpha-\beta|}{2},\gamma\}= \frac{\frac{\alpha+\beta-|\alpha-\beta|}{2}+\gamma-|\frac{\alpha+\beta-|\alpha-\beta|}{2}-\gamma|}{2}$$
So by continuity we have the answer.  
