Prove that $x_{n}$ identified by : with all $n\in \mathbb{N}, x_{n}= Sup(u_{n},v_{n})$ converges. Let $u_{n},v_{n}$ are real serial numbers and converge. Prove that $x_{n}, y_{n}$ identified by : with all $n\in \mathbb{N}, \left\{\begin{matrix}
x_{n}= &Sup(u_{n},v_{n}) \\ 
y_{n}= &Inf(u_{n},v_{n}) 
\end{matrix}\right.$ converge.

I firstly approach to this field, so there are many thing make me confused. Please help me. Thank you so much!
Firstly, let $u_{n},v_{n}$ converge to $a,b$ respectively, and $b<a$. We solve for $x_{n}$.
I think with $n, N\in \mathbb{N}$ then we can choose $u_{N},v_{N}$ satisfying: $Sup(u_{N},v_{N})=\frac{1}{2}(u_{N}+v_{N}+|u_{N}-v_{N}|)$

*

*However I wonder that if we can only choose $u_{N}$ from an interval, does $x_{N}=Sup(u_{N},v_{N})$ exist?


*Another problem is that if the question above is solved, can I perform my solution as this or I need to find exactly $\varepsilon$ is?:
"$\forall \varepsilon >0,\exists N\in \mathbb{N}, \forall n\in \mathbb{N}, (n\geq N\Rightarrow |x_{n}-x_{N}|\leq \varepsilon )$"

*

*In case we need to find $lim x_{n}$, then will $a$ be $lim x_{n}$?

 A: I split up the proof for $(x_n)$ converges into 2 cases. One where $a=b$ and the other where $a \neq b$.
Suppose $a=b$
Choose some $\epsilon >0.$
Then we know that $\exists N_1,N_2$ s.t. $\forall n>N_1 , m>N_2,$
$$
\vert u_n-a\vert < \epsilon \quad \textrm{and} \quad \vert v_m -b\vert < \epsilon.
$$
Let $M = max\{N_1 , N_2\}.$ Then, $\forall q > M,$
$$\tag{1}
\vert u_q-a\vert < \epsilon \quad \textrm{and} \quad \vert v_q -b\vert < \epsilon.
$$
Therefore, $(1)$ holds for $sup(u_q,v_q)$. Thus, $(x_n)$ converges.
Suppose $a \neq b$
Wlog, assume $b<a$ and let $\epsilon >0.$
Then we know that $\exists N_1,N_2$ s.t. $\forall n>N_1 , m>N_2,$
$$\tag{2}
\vert u_n-a\vert < \epsilon \quad \textrm{and} \quad \vert v_m -b\vert < \epsilon.
$$
Let $M = max\{N_1 , N_2 \}$ and let $0<\epsilon < \frac{a-b}{3}$.
Then, using the above and $(2)$ we get $\forall q > M$
$$
\vert u_q-a\vert < \epsilon < \frac{a-b}{3}  \quad \textrm{and} \quad \vert v_q -b\vert < \epsilon < \frac{a-b}{3}
$$
Which implies, $x_q = sup(u_q,v_q) = u_q$ $\tag{3}$
Thus, $(x_n)$ converges to $a$.
$Q.E.D.$
Where $(3)$ holds as $b<a$ and we are now only considering the intervals $\big[b - \frac{a-b}{3},b+ \frac{a-b}{3}\big], \big[a- \frac{a-b}{3},a +\frac{a-b}{3} \big]$. Consider drawing a diagram if you don't seem to follow.
The proof for $(y_n)$ is similar to the proof above but uses $inf$.
A: Regarding you first question:
As we are considering the real numbers we can make use of the supremum axiom which states that every bounded set of real numbers has a unique supremum. Both sequences $(u_n)$ and $(v_n)$ are convergent and hence bounded so $x_n=\sup(u_n,v_n)$ exists for all $n\in\mathbb{N}$.
I would solve the question as follows:
Let be $\epsilon:=a-b>0$ then due to convergence of $(u_n)$ and $(v_n)$ I can find two indices $N_u$ and $N_v$ such that for all members of $(u_n)$ with $n> N_u$ it holds that $|u_n-b|<\frac{\epsilon}{2}$ and for all members of $(v_n)$ with $n> N_v$ it holds that $|v_n-a|<\frac{\epsilon}{2}$. Let be $N:=\max\{N_u,N_v\}$ then for all $n>N$ we see that  $u_n$ and $v_n$ satisfy $u_n<b+\frac{\epsilon}{2}=a-\frac{\epsilon}{2}<v_n$. So we can conclude that the supremum $x_n=\sup(u_n,v_n)$ exists as $(u_n,v_n)$ is bounded for all $n\in\mathbb{N}$. Furhter, it is always a member of the sequence $(v_n)$ or more precisely $x_n= v_n$ for all $n>N$. As the sequence $v_n$ converges it follows that the sequence $(x_n)$ converges as well.
The case where you consider the sequence $(y_n)$ is handled in the same way - just replace the $\sup$ with the $\inf$.
