1. ## Real Anaylsis help

Ok, So I am having some trouble with my real analysis class and homework, and was wanting some help...

1) Show that if (x_n) and (y_n) are convergent sequences, then the sequences (u_n) and (v_n) defined by (u_n):=max{ x_n , y_n) and (v_n):=min{ x_n , y_n } are also convergent.

2. Originally Posted by osudude
1) Show that if (x_n) and (y_n) are convergent sequences, then the sequences (u_n) and (v_n) defined by (u_n):=max{ x_n , y_n) and (v_n):=min{ x_n , y_n } are also convergent.
Suppose that $\displaystyle \left( {x_n } \right) \to X\;\& \;\left( {y_n } \right) \to Y$.
There are to cases to consider: $\displaystyle X = Y\;\& \;X \ne Y$.

Here a hint on the second one.
Suppose that (WLOG) $\displaystyle Y < X$ then $\displaystyle Z=\frac{X+Y}{2}$ so that $\displaystyle Y<Z<X$.
Thus almost all of the terms of $\displaystyle \left( {x_n } \right)$ are greater than $\displaystyle Z$.

And almost all of the terms of $\displaystyle \left( {y_n } \right)$ are less than $\displaystyle Z$.

What does that tell you about $\displaystyle \max\text{ and }\min?$

3. that they coverge both to z since the max of y is almost z and same for the min of x??

4. ## Relation between max(min) and absolute value

In fact , we have $\displaystyle u_n=max\{x_n,y_n\}=\frac {1}{2}(x_n+y_n+|x_n-y_n |)$, similarly $\displaystyle v_n=min \{x_n,y_n\}=\frac {1}{2}(x_n+y_n-|x_n-y_n |)$.
since $\displaystyle x_n$ and $\displaystyle y_n$ converges, so is $\displaystyle u_n$ and $\displaystyle v_n$.