TIFR 2014 Problem 29 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 programme 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.
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Let $f: X\to Y$ be a continuous map between metric spaces. Then $f(X)$ is a complete subset of $Y$ if
A. X is compact
B. Y is compact
C. X is complete
D. Y is complete
Discussion:
Let $X$ be compact. Then $f(X)$ is compact. (continuous image of compact space is compact)
Now, compact subset of any Hausdorff space is closed. So in particular, compact subset of any metric space is closed.
Let $y_n$ be a Cauchy sequence in $f(X)$. Then since $f(X)$ is compact, $y_n$ has a convergent subsequence (converging to a point in that compact set i.e, in $f(X)$.
Suppose $y_{n_k} \to y \in f(X)$.
Then by triangle inequality, we have $d(y_n,y) \le d(y_n,y_{n_k}) + d(y_{n_k},y) \to 0+0=0$ as $n\to \infty$
Here we have used that $y_n$ is cauchy to conclude $d(y_n,y_{n_k}) \to 0$.
So this implies $y_n \to y$. Since $y\in f(X)$ we conclude that $f(X)$ is complete.
This proves A.
Let $Y=[0,2]$ and $X=(0,1)$. Take the inclusion map $i(x)=x$ for all $x\in X$. This example shows that even if we take $Y$ to be compact, or complete, $f(X)$ need not be complete. So this disproves B and D.
Now take $X=\mathbb{R}$ and take $Y=(0,1)$. We know there is a homeomorphism between these two sets where the metric is usual topology. So, in this case, the image of a complete set is not complete. This disproves option C.
