 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.
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## Problem:

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.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Cauchy Sequence,Homeomorphishm, Compact Space
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert