 TIFR 2014 Problem 15 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:

$$X$$ is a metric space. $$Y$$ is a closed subset of $$X$$ such that the distance between any two points in $$Y$$ is at most 1. Then

A. $$Y$$ is compact.

B. any continuous function from $$Y\to \mathbb{R}$$ is bounded.

C. $$Y$$ is not an open subset of $$X$$

D. none of the above.

Discussion:

Let $$X=$$ an infinite set for example $$=\mathbb{R}$$ with the metric as discrete metric.

That is $$d(x,y)=1$$ if $$x\neq y$$ and $$d(x,y)=0$$ if $$x=y$$.

Then every set in $$X$$ is open and every set is closed.

Now take $$Y=X$$. Then $$Y$$ can be covered by singleton sets. $$Y=\cup \{\{a\}|a\in Y\}$$. Now each of the singleton sets is open in discrete metric space. Therefore, this is an open cover for $$Y$$. Since $$Y$$ is infinite, this cover has no finite subcover. So $$Y$$ is not compact.

Given any $$f:Y\to \mathbb{R}$$, for open set $$U\in \mathbb{R}$$, $$f^{-1}(U)\subset Y$$. Since Y is discrete, $$f^{-1}(U)$$is open in $$Y$$. So every function $$f:Y\to \mathbb{R}$$ is a continuous function. In particular if we define $$f(x)=x$$ then $$f$$ is a continuous function. And $$f$$ is not bounded.

Also, every subset of $$X$$ is open. So $$Y$$ is open.

Therefore, we are left with none of the above.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous function, Discrete Metric Space, Finite subcover
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert