Cheenta

Academy for Gifted Students

How Cheenta works to ensure student success?

Explore the Back-Story

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

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

**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.

**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

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

**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.

**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

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Online Live Classroom Programs

Online Self Paced Programs [*New]

Past Papers

More