Real Anaylsis help

**osudude** · November 12th 2009, 11:00 AM

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.

**Plato** · November 12th 2009, 11:45 AM

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 $\left( {x_n } \right) \to X\;\& \;\left( {y_n } \right) \to Y$ .
There are to cases to consider: $X = Y\;\& \;X \ne Y$ .

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

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

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

**osudude** · November 14th 2009, 08:43 AM

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

**Shanks** · November 18th 2009, 05:04 AM

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

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