Assignment # 01 MTH621 (Spring 2022) Due Date: June 07, 2022

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Assignment # 01                         MTH621 (Spring 2022)

 

Maximum Marks:  20                                                                                   Due Date: June 07, 2022

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Q1.

      Prove or disprove that the complement of integers {Z^c} in R is an open set.

SOLUTION:-

      A set U \subset R is open if and only if for every x \in U , there exists some \in > 0 such that \left( {x - \in ,x + \in } \right) is a subset of U.

For U = Z this is clearly not the case:

Take

       x = 0

      Take any  

               { \in > 0} 

Then,

          \min \left\{ {x + \frac{ \in }{2},x + \frac{1}{2}} \right\} is an element of \left( {x - \in ,x + \in } \right), but it is not an element of Z.

Therefore, 

\left( {x - \in ,x + \in } \right) is not a subset of Z for any value of  \in.

Therefore,

Z is not open.

If Z is not open then {Z^c} is open.

Hence,

             Prove that the complement of integers {Z^c} in R is an open set.

Q2.

Show that the sequence \left\{ {\sqrt {n + 1} - \sqrt n } \right\}_{n = 1}^\infty  is a null sequence.

Solution:-

Consider the function f(n) defined by

        f\left( n \right) = \sqrt {n + 1} - \sqrt n

First, we ues multiply by the numerator and denominator f\left( n \right) = \sqrt {n + 1} + \sqrt n.

         f\left( n \right) = \sqrt {n + 1} - \sqrt n \times \frac{{\sqrt {n + 1} + \sqrt n }}{{\sqrt {n + 1} + \sqrt n }}

        f\left( n \right) = \frac{{\left( {\sqrt {n + 1} - \sqrt n } \right)\left( {\sqrt {n + 1} + \sqrt n } \right)}}{{\sqrt {n + 1} + \sqrt n }}

        f\left( n \right) = \frac{{{{\left( {\sqrt {n + 1} } \right)}^2} - {{\left( {\sqrt n } \right)}^2}}}{{\sqrt {n + 1} + \sqrt n }}

          f\left( n \right) = \frac{{n + 1 - n}}{{\sqrt {n + 1} + \sqrt n }}

          f\left( n \right) = \frac{1}{{2\sqrt n }}

Then,

        Since \sqrt {n + 1} > \sqrt n we have.

         1 < \frac{1}{{\sqrt {n + 1} + \sqrt n }} < \frac{1}{{\sqrt n + \sqrt n }}

         1 < \frac{1}{{\sqrt {n + 1} + \sqrt n }} < \frac{1}{{2\sqrt n }}

Since, 

   We know from a property of Apostol that:

{\lim }\limits_{n \to \infty } \frac{1}{{{n^\alpha }}} = 0,\,\,\,\,\,\,\,\,\,\left( {\alpha > 0} \right)

 {\lim }\limits_{n \to \infty } \frac{1}{{2\sqrt n }} = 0

{\lim }\limits_{n \to \infty } f\left( n \right) = 0

Prove that:-

             The sequence \left\{ {\sqrt {n + 1} - \sqrt n } \right\}_{n = 1}^\infty is a null sequence.


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