TIFR 2015 Problem 7 Solution -Increasing Function and Continuity

TIFR 2015 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate program leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let $f$ and $g$ be two functions from $([0,1])$ to $([0,1])$ with $f$ strictly increasing. Which of the following statements is always correct?

A. If (g) is continuous, then $(f\circ g)$ is continuous

B. If (f) is continuous, then $(f \circ g)$ is continuous

C. If (f) and $(f \circ g)$ is continuous, then (g) is continuous

D. If (g) and $(f \circ g)$ are continuous, then (f) is continuous

Discussion:

A: Let (g(x)=x) for all $(x\in [0,1])$ .

(f(x)=x) for $(x \in [0,\frac{1}{2}])$ and (f(x)=5+x) for $(x\in (\frac{1}{2},1])$ .

Then $(f \circ g=f)$ and (f) is not continuous.

So A is False.

B: Reverse (f) and (g) in A to show that B is False.

C: If (f) and $(f \circ g)$ are continuous then (f) is 1-1 (increasing), continuous map ([0,1]to [0,1]).

(A subset [0,1] ) be closed. Then (A) is compact. (Closed subsets of compact spaces are compact).

Therefore (f(A)) is compact. (continuous image of compact set is compact).

We have that (f(A)) is a compact subset of ([0,1]). Therefore (f(A)) is closed in ([0,1]). (compact subspace of Hausdorff space is closed).

Therefore, (f) is a closed map. So $(f^{-1})$ is continuous.

Hence $(f^{-1} \circ f \circ g=g)$ is continuous.

So, C is True.

D: Let $(g(x)=\frac{x}{4})$ for all $(x\in [0,1])$ .

(f(x)=x) for $(x \in [0,\frac{1}{2}])$ and (f(x)=5+x) for $(x\in (\frac{1}{2},1])$ .

Then $(f \circ g(x)=f(\frac{x}{4})=\frac{x}{4})$ for all $(x \in [0,1])$ .

So $(f \circ g)$ is continuous but (f) is not continuous.

So, D is False.

Helpdesk

What is this topic: Real Analysis
What are some of the associated concept: Continuity,Closed Set, Compact Set
Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

TIFR 2015 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate program leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let $f$ and $g$ be two functions from $([0,1])$ to $([0,1])$ with $f$ strictly increasing. Which of the following statements is always correct?

A. If (g) is continuous, then $(f\circ g)$ is continuous

B. If (f) is continuous, then $(f \circ g)$ is continuous

C. If (f) and $(f \circ g)$ is continuous, then (g) is continuous

D. If (g) and $(f \circ g)$ are continuous, then (f) is continuous

Discussion:

A: Let (g(x)=x) for all $(x\in [0,1])$ .

(f(x)=x) for $(x \in [0,\frac{1}{2}])$ and (f(x)=5+x) for $(x\in (\frac{1}{2},1])$ .

Then $(f \circ g=f)$ and (f) is not continuous.

So A is False.

B: Reverse (f) and (g) in A to show that B is False.

C: If (f) and $(f \circ g)$ are continuous then (f) is 1-1 (increasing), continuous map ([0,1]to [0,1]).

(A subset [0,1] ) be closed. Then (A) is compact. (Closed subsets of compact spaces are compact).

Therefore (f(A)) is compact. (continuous image of compact set is compact).

We have that (f(A)) is a compact subset of ([0,1]). Therefore (f(A)) is closed in ([0,1]). (compact subspace of Hausdorff space is closed).

Therefore, (f) is a closed map. So $(f^{-1})$ is continuous.

Hence $(f^{-1} \circ f \circ g=g)$ is continuous.

So, C is True.

D: Let $(g(x)=\frac{x}{4})$ for all $(x\in [0,1])$ .

(f(x)=x) for $(x \in [0,\frac{1}{2}])$ and (f(x)=5+x) for $(x\in (\frac{1}{2},1])$ .

Then $(f \circ g(x)=f(\frac{x}{4})=\frac{x}{4})$ for all $(x \in [0,1])$ .

So $(f \circ g)$ is continuous but (f) is not continuous.

So, D is False.

Helpdesk

What is this topic: Real Analysis
What are some of the associated concept: Continuity,Closed Set, Compact Set
Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert