MTH634: Topology Assignment No. 1 Spring 2022 Due Date: Monday, June 27, 2022

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Spring 2022                 MTH634: Topology

Assignment No. 1 (Lectures No. 16 to 20-Topic#51 to 76)        

Maximum Marks: 10

Due Date: Monday, June 27, 2022

 

Please read the following instructions carefully before attempting the solution of this assignment:

(1)   To solve this assignment, you should have good command over Topic# 51 to 76 (Lecture No. 16 to 20).

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Question:                                                                                                              Marks: 10

 

Show that the functions, f\left( x \right) = 2x + 3, g\left( x \right) = \frac{x}{2}and h\left( x \right) = {x^2} are all continuous functions mapping \mathbb{R} to \mathbb{R} in the usual topology with the basis element, (a, b)where a<b.

 

Solution:                                     

 

Type your solution here

Definition

Continuous functions:

                                     A function f:X \to Y is said to be continuous if the inverse image of every open subset of Y is open in X. In other words, if v \in {\Im _Y} then its inverse image {f^{ - 1}}\left( V \right) \in {\Im _X}.

So, given that:

         f\left( x \right) = 2x + 3

Let x = 1 and 2

        f\left( x \right) = 2x + 3

Taking put the value x = 1 in above eq.

        \begin{array}{l} f\left( 1 \right) = 2(1) + 3\\ f\left( 1 \right) = 2 + 3\\ f\left( 1 \right) = 5 \end{array} 
f\left( x \right) = 2x + 3

Taking put the value x = 2 above eq.

\begin{array}{l} f\left( 2 \right) = 2(2) + 3\\ f\left( 2 \right) = 4 + 3\\ f\left( 2 \right) = 7 \end{array}

f\left( 1 \right) < f\left( 2 \right) is increasing function.

Now,

         \begin{array}{l} f(x) = y\\ x = {f^{ - 1}}(y)\\ 2x + 3 = y\\ 2x = y - 3 \end{array}

       \begin{array}{l} x = \frac{{y - 3}}{2}\\ {f^{ - 1}}(y) = \frac{{y - 3}}{2}\\ {f^{ - 1}}(x) = \frac{{x - 3}}{2} \end{array}

Inverse image of every open subset of Y is open in X.

Open interval of function \left( {\frac{{a - 3}}{2},\frac{{b - 3}}{2}} \right).

Now, given that:

g\left( x \right) = \frac{x}{2}

Let x = 1 and 2

g\left( x \right) = \frac{x}{2}

Taking put the value x = 1 above eq.

\begin{array}{l} g\left( x \right) = \frac{x}{2}\\ g\left( 1 \right) = \frac{1}{2} \end{array}

g\left( x \right) = \frac{x}{2}

Taking put the value x = 2 above eq.

\begin{array}{l} g\left( x \right) = \frac{x}{2}\\ g\left( 2 \right) = \frac{2}{2}\\ g\left( 2 \right) = 1 \end{array}

g\left( 1 \right) < g\left( 2 \right) is increasing function.

Now,

\begin{array}{l} g\left( x \right) = y\\ x = {g^{ - 1}}(y)\\ \frac{x}{2} = y \end{array}

\begin{array}{l} x = 2y\\ {g^{ - 1}}(y) = 2y\\ {g^{ - 1}}(x) = 2x \end{array}

Inverse image of every open subset of Y is open in X.

Open interval of function (2a, 2b).

Now, given that:

h\left( x \right) = {x^2}

Let x = 1 and 2

h\left( x \right) = {x^2}

\begin{array}{l} h\left( x \right) = {x^2}\\ h\left( 1 \right) = {(1)^2}\\ h\left( 1 \right) = 1 \end{array}

h\left( x \right) = {x^2}

Taking put the value x = 2 above eq.

\begin{array}{l} h\left( x \right) = {x^2}\\ h\left( 2 \right) = {(2)^2}\\ h\left( 2 \right) = 4 \end{array}

h(1) < h(2) 

is increasing function.

Now,

\begin{array}{l} h(x) = y\\ x = {h^{ - 1}}(y)\\ {x^2} = y \end{array}

Taking both sides square root.

\begin{array}{l} \sqrt {{x^2}} = \pm \sqrt y \\ x = \pm \sqrt y \\ {h^{ - 1}}(y) = \pm \sqrt y \\ {h^{ - 1}}(x) = \pm \sqrt x \end{array}

Inverse image of every open subset of Y is open in X.

Open interval of function:

{h^{ - 1}}(x) =\left\{ {\left( { - \sqrt a , - \sqrt b } \right) \cup \left( {\sqrt a ,\sqrt b } \right)\,\,\,\,\,\,\,\,\,\,if\,0 \le a < b} \right.

                 \left\{ {\left( { - \sqrt b ,\sqrt b } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,a < 0 < b\,} \right.

                 \left\{ {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,b < 0\,} \right.

So, hence prove that:

The functions f(x) = 2x + 3, g(x) = \frac{x}{2} and are all continuous functions mapping \mathbb{R} to \mathbb{R} in the usual topology with the basis element, (a, b) where a<b.

                                  

 





 

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